Question 1

Evaluate the expression 1 / log₃ 60 + 1 / log₄ 60 + 1 / log₅ 60, where all logarithms are to their indicated bases.

Accepted Answer

Introduction / Context: This question explores a useful identity involving reciprocals of logarithms with different bases but the same argument. The expression consists of three terms: 1 / log₃ 60, 1 / log₄ 60, and 1 / log₅ 60. By recognising how reciprocals of logs convert into logs with a common base, we can simplify the sum and show that it reduces to a very simple value. Given Data / Assumptions: We need to evaluate S = 1 / log₃ 60 + 1 / log₄ 60 + 1 / log₅ 60. Here log₃ 60 means log of 60 to base 3, and similarly for bases 4 and 5. All logs are real and defined for positive arguments and bases not equal to 1. We may use the identity 1 / logₐ b = log_b a. Concept / Approach: The key identity is 1 / logₐ b = log_b a. This follows from the change of base formula. Applying this identity to each term in the sum will convert all the terms into logs with base 60. Specifically, 1 / log₃ 60 becomes log₆₀ 3, and similarly for the other terms. Then we can use properties of logarithms to combine the sum into a single log expression whose value is easy to recognise. Step-by-Step Solution: Step 1: Start with S = 1 / log₃ 60 + 1 / log₄ 60 + 1 / log₅ 60. Step 2: Use the identity 1 / logₐ b = log_b a for each term. Thus 1 / log₃ 60 = log₆₀ 3, 1 / log₄ 60 = log₆₀ 4, and 1 / log₅ 60 = log₆₀ 5. Step 3: Substitute to obtain S = log₆₀ 3 + log₆₀ 4 + log₆₀ 5. Step 4: Use the log property logₖ a + logₖ b = logₖ (a b) with common base 60. Combine the first two terms: log₆₀ 3 + log₆₀ 4 = log₆₀ (3 × 4) = log₆₀ 12. Step 5: Add the third term: log₆₀ 12 + log₆₀ 5 = log₆₀ (12 × 5) = log₆₀ 60. Step 6: Since log₆₀ 60 equals 1, the sum S equals 1. Verification / Alternative check: The identity 1 / logₐ b = log_b a can be re derived from the change of base formula. Using that, we have log₆₀ 3 + log₆₀ 4 + log₆₀ 5, which describes the log of the product 3 × 4 × 5 using base 60. The product is exactly 60, so log₆₀ 60 must be 1 by definition, confirming our result without needing numerical approximations. Why Other Options Are Wrong: Options 0, 5, and 60 are inconsistent with the identities used. There is no cancellation that would produce 0, and the product 3 × 4 × 5 is 60, not 1 or 60 squared. So the combined log cannot reasonably give 5 or 60. The sum simplifies precisely to log₆₀ 60, which is 1, making option 1 the only correct choice. Common Pitfalls: Students may try to compute each log numerically, which is unnecessary and error prone. Others might forget that the reciprocal identity changes both base and argument. Another frequent mistake is to combine logs with different bases directly, which is not allowed. The correct approach always converts them to a common base before combining. Final Answer: The value of 1 / log₃ 60 + 1 / log₄ 60 + 1 / log₅ 60 is 1.

Question 2

If log₈ x = 3/13 (logarithm to base 8), then what is the value of x? Choose the correct option from the following.

Accepted Answer

Introduction / Context: This question involves solving a logarithmic equation for the unknown argument. We are given log₈ x = 3/13 and must find x. Instead of expecting a simple integer result, we must carefully apply the exponential form of the logarithm definition. The options include several integers and a general option None of these, so a key part of the solution is to check whether the derived value matches any of the integers given. Given Data / Assumptions: The equation is log₈ x = 3/13. We require x to be positive, since logarithms of non positive numbers are not defined in the real system. The base 8 is positive and not equal to 1. We must choose the correct x among 25, 32, 37, 16 or None of these. Concept / Approach: From the definition of logarithm, log₈ x = 3/13 means 8^(3/13) = x. Thus we can express x directly as a fractional power of 8. By simplifying 8 as 2³ and carefully handling exponents, we can express x in terms of a power of 2 and then compare its size with the integer options. This will show whether x equals any of the proposed integers or not. Step-by-Step Solution: Step 1: Use the definition of logarithm. From log₈ x = 3/13, we have x = 8^(3/13). Step 2: Write 8 in terms of powers of 2: 8 = 2³. Step 3: Substitute into the expression for x: x = (2³)^(3/13). Step 4: Use the power of a power rule: (aᵐ)ⁿ = a^(m n). Here, x = 2^(3 × 3/13) = 2^(9/13). Step 5: The exponent 9/13 is a positive fraction less than 1, so x is greater than 1 but less than 2, because 2¹ is 2 and 2⁰ is 1. Step 6: Therefore x is a positive real number between 1 and 2 and clearly cannot be equal to 25, 32, 37, or 16, all of which are much larger than 2. Verification / Alternative check: We can approximate x numerically. Using a calculator or estimation, 8^(3/13) is roughly 1.616. Check against options: 25, 32, 37, and 16 are all much larger than 1.616. Since none of the given integers is close to this value, the correct answer cannot be any of these integers; the only remaining option is None of these. Why Other Options Are Wrong: If we suppose x = 25, then log₈ 25 would be roughly log 25 divided by log 8, which is not 3/13. Similarly, x = 32 would require 8^(3/13) to equal 32, but we know 8² = 64 and 8¹ = 8, so there is no exponent as small as 3/13 giving 32. The same reasoning eliminates 37 and 16. Each of these values gives a very different logarithm than 3/13 when checked directly. Common Pitfalls: Some learners guess that x must be a simple integer like 32 or 16 because powers of 2 are common, but they overlook that the exponent 3/13 is a small fraction. Others mistakenly solve 8^(13/3) instead of 8^(3/13). Carefully applying the rule logₐ b = c implies aᶜ = b is crucial for avoiding such errors. Final Answer: The correct choice is None of these, because x = 8^(3/13) = 2^(9/13), which is not equal to any listed integer option.

Question 3

Using log10 3 = 0.47712, find the number of digits in the decimal representation of 3 raised to the power 33.

Accepted Answer

Introduction / Context: This question tests the standard logarithm technique for finding how many decimal digits are in a very large power such as 3^33. Direct computation of 3^33 is not practical in an exam setting, so students must know the connection between the common logarithm log10 N and the number of digits in N. Such questions appear frequently in aptitude exams and engineering entrance tests, and they are a good check of both logarithm rules and basic understanding of place value in the decimal system. Given Data / Assumptions: - We are given log10 3 = 0.47712. - We must find the number of digits in 3^33 when written in base 10. - The number 3^33 is positive, so its common logarithm is defined. - All logarithms are taken to base 10 unless otherwise specified. Concept / Approach: If N is a positive integer, then the number of decimal digits in N is given by digits = floor(log10 N) + 1. This follows from the fact that 10^(k) <= N < 10^(k+1) means N has k + 1 digits. For a power like 3^33 we first compute log10(3^33) using the power rule log10(a^n) = n * log10 a, and then apply the digit formula. The key idea is to avoid computing 3^33 directly and instead work entirely with logarithms. Step-by-Step Solution: Let N = 3^33. Compute log10 N using the power rule: log10 N = log10(3^33) = 33 * log10 3. Substitute the given value log10 3 = 0.47712. So log10 N = 33 * 0.47712. Multiply: 33 * 0.47712 = 15.74496. Therefore log10 N = 15.74496. The number of digits in N is floor(15.74496) + 1. floor(15.74496) = 15, so digits = 15 + 1 = 16. Verification / Alternative check: We know 10^15 has 16 digits and 10^16 has 17 digits. The value 3^33 has a common logarithm of approximately 15.74496, which lies between 15 and 16. That means 10^15 < 3^33 < 10^16, so 3^33 must have more digits than 10^15 but fewer digits than 10^16, giving exactly 16 digits. This confirms the result obtained from the floor and plus one formula. Why Other Options Are Wrong: 14: This would correspond to log10(3^33) being between 13 and 14, which is much smaller than the computed value 15.74496. 15: This would require log10(3^33) to be between 14 and 15, which again does not match 15.74496. 17: This would require log10(3^33) to be at least 16, which is larger than 15.74496, so 17 digits is impossible. Common Pitfalls: A common mistake is to forget the plus one in the digit formula and answer floor(log10 N) instead of floor(log10 N) + 1. Another frequent error is to round 15.74496 to 16 and incorrectly use that rounded value directly as the number of digits. Students may also misapply logarithm rules, for example by writing log10(3^33) as (log10 3)^33, which is incorrect. Careful use of the power rule and the exact digit formula avoids these errors. Final Answer: The number of digits in 3^33 is 16.

Question 4

If log base a of b plus log base b of a is equal to 2, that is log_a b + log_b a = 2, then which of the following relations between a and b must be true?

Accepted Answer

Introduction / Context: This question checks understanding of logarithms with variable bases and arguments. In particular it tests the reciprocal relationship between log_a b and log_b a, and how to convert a logarithmic equation into a simpler algebraic equation. Such problems are very common in aptitude and algebra tests and they help students become comfortable working with logs whose base is not a fixed number like 10 or e. Given Data / Assumptions: - log_a b + log_b a = 2. - a and b are positive real numbers and neither a nor b is equal to 1, so all logarithms are defined. - We must find a relation between a and b that always holds under this condition. Concept / Approach: The key identity is that log_b a is the reciprocal of log_a b. More precisely, for positive a and b with a not equal to 1 and b not equal to 1, we have log_a b = 1 / log_b a. We use this to convert the given expression into a single variable equation. Once we reduce it to a standard quadratic equation, we solve it and interpret the result back in terms of a and b. Step-by-Step Solution: Let x = log_a b. Then, by the change of base property, log_b a = 1 / x. The given condition becomes x + 1 / x = 2. Multiply through by x to clear the denominator: x^2 + 1 = 2x. Rearrange to obtain a quadratic: x^2 - 2x + 1 = 0. Factor: (x - 1)^2 = 0, so x = 1. Thus log_a b = 1, which means b = a^1 = a. Therefore the relation that must hold is a = b. Verification / Alternative check: If a and b are equal, say a = b = k with k greater than 0 and not equal to 1, then log_a b = log_k k = 1 and log_b a = 1. Their sum is 1 + 1 = 2, which satisfies the given equation. If a is not equal to b, then log_a b is not 1, and the reciprocal sum x + 1 / x can never be equal to 2 unless x = 1. This confirms that a = b is the only possible relation. Why Other Options Are Wrong: a + b = 1: This is not forced by the logarithmic equation and may even fail for positive numbers greater than 1. a - b = 1: This would only hold for very specific pairs, but the logarithmic equation demands a general relation that works for all valid a and b. ab = 1: This also does not follow from the derived condition x = 1 and is unrelated to the sum of logs being 2. Common Pitfalls: One common mistake is forgetting that log_a b and log_b a are reciprocals, not equal values. Students might also incorrectly apply change of base and treat log_a b as log b / log a without keeping track of the base. Another error is failing to handle the quadratic equation correctly, for example by discarding the unique solution x = 1. Remembering that x + 1 / x has minimum value 2 when x is 1 also provides a useful shortcut check. Final Answer: The only relation that always holds under the given condition is a = b.

Question 5

For x greater than 1, find the least possible value of the expression 2 log10 x - log_x (1 / 100).

Accepted Answer

Introduction / Context: This question tests the ability to work with logarithms that have different bases and to analyze a function in order to find its minimum value for x greater than 1. To solve it efficiently, you must use the change of base formula for logarithms and then apply basic calculus or inequality ideas to minimize a simple one variable expression. It is a standard type of aptitude question that mixes algebraic manipulation with an understanding of how logarithmic functions behave. Given Data / Assumptions: - The variable x satisfies x greater than 1. - The expression to be minimized is E(x) = 2 log10 x - log_x (1 / 100). - All logarithms are common logarithms (base 10) unless another base is explicitly written. - The expression is defined for x greater than 1 because log_x (1 / 100) requires x positive and not equal to 1. Concept / Approach: We first convert the term log_x (1 / 100) into a base 10 logarithm using the change of base formula log_x y = log10 y / log10 x. Then we express the entire function E(x) in terms of t = log10 x, which is positive for x greater than 1. This converts the problem into minimizing a simple algebraic expression in t of the form 2t + 2 / t. That expression can be minimized using basic derivative ideas or the inequality between arithmetic and geometric means. Step-by-Step Solution: Let E(x) = 2 log10 x - log_x (1 / 100). Use change of base: log_x (1 / 100) = log10 (1 / 100) / log10 x. Note that 1 / 100 = 10^-2, so log10 (1 / 100) = -2. Therefore log_x (1 / 100) = -2 / log10 x. So E(x) = 2 log10 x - ( -2 / log10 x ) = 2 log10 x + 2 / log10 x. Let t = log10 x. Since x greater than 1, t is greater than 0. Then E(t) = 2t + 2 / t for t greater than 0. Differentiate with respect to t: E'(t) can be used to confirm a minimum but first find where E'(t) is positive. Compute the first derivative: E'(t) is not needed; instead, dE/dt = 2 - 2 / t^2. Set dE/dt = 0: 2 - 2 / t^2 = 0 implies 2 = 2 / t^2, so t^2 = 1. Because t is greater than 0, t = 1. Substitute back: E_min = 2 * 1 + 2 / 1 = 2 + 2 = 4. Verification / Alternative check: We can also use the arithmetic mean and geometric mean inequality. For t greater than 0, the arithmetic mean of the two positive numbers 2t and 2 / t is at least twice the geometric mean. That is, (2t + 2 / t) / 2 is greater than or equal to sqrt( (2t) * (2 / t) ) = 2. Hence 2t + 2 / t is greater than or equal to 4, and equality holds when 2t = 2 / t, so t^2 = 1 and t = 1. This confirms that the least value of the expression is 4. Why Other Options Are Wrong: 2: This would violate the inequality 2t + 2 / t greater than or equal to 4 for t greater than 0. 3: This is less than 4 and therefore cannot be the minimum value. 5: This is larger than the minimum 4; it is simply another attainable value for some other x, not the least value. Common Pitfalls: A common error is to forget that log_x (1 / 100) is negative and to write it as 2 / log10 x instead of minus 2 / log10 x. Another mistake is to try to minimize the original expression directly in terms of x, which is more complex than working with t = log10 x. Some students also confuse the formula for the number of digits with this minimization problem, because both use log10. Carefully converting to a single variable and using either differentiation or the arithmetic mean and geometric mean inequality avoids these problems. Final Answer: The least possible value of the given expression is 4.

Question 6

Solve the equation 12^(2x + 1) = 1 and find the real value of x.

Accepted Answer

Introduction / Context: This question is a simple exponential equation that checks whether students understand how to solve equations of the form a^(f(x)) = 1. It also connects the behaviour of exponential functions with their exponents. Such problems are common in aptitude and entrance tests because they combine algebra with basic properties of exponents and do not require heavy computation. Given Data / Assumptions: - The equation is 12^(2x + 1) = 1. - The base 12 is a positive real number and not equal to 1. - We are looking for real solutions for x. Concept / Approach: For an exponential expression of the form a^k with a greater than 0 and a not equal to 1, the value a^k equals 1 if and only if the exponent k equals 0. This is because a^0 = 1 for any such base a, and there is no other real exponent that gives 1 when the base is not 1. Therefore we can solve this equation simply by setting the exponent 2x + 1 equal to 0 and solving the resulting linear equation. Step-by-Step Solution: We are given 12^(2x + 1) = 1. Note that 12 is greater than 0 and not equal to 1. For such a base, a^k = 1 implies k = 0. Therefore set the exponent equal to zero: 2x + 1 = 0. Solve the linear equation: 2x = -1. Then x = -1 / 2. Thus the unique real solution of the original equation is x = -1 / 2. Verification / Alternative check: Substitute x = -1 / 2 back into the original equation. The exponent becomes 2(-1 / 2) + 1 = -1 + 1 = 0. Therefore the left hand side is 12^0 = 1. This matches the right hand side exactly, confirming that x = -1 / 2 is correct. Because the exponential function 12^(2x + 1) is continuous and strictly monotonic in x, there can be no other real solution to this equation. Why Other Options Are Wrong: 1 / 2: Substituting x = 1 / 2 gives exponent 2(1 / 2) + 1 = 2, so 12^2 which is 144, not 1. 1: Substituting x = 1 gives exponent 2(1) + 1 = 3, so 12^3 which is 1728, not 1. -1: Substituting x = -1 gives exponent 2(-1) + 1 = -1, so 12^-1 which is 1 / 12, not 1. Common Pitfalls: Some students mistakenly think that any exponent can give 1 if the base is large, which is not correct. Others confuse this situation with a^k = 0 or a^k = a and try to set the exponent to 1 instead of 0. Another error is to attempt to take logarithms on both sides without realizing that the simple property a^0 = 1 makes the problem trivial. Remember that for a base a not equal to 1, the only way to get a^k = 1 is to have k equal to 0. Final Answer: The solution of the equation 12^(2x + 1) = 1 is x = -1 / 2.

Question 7

For x in the natural numbers with x greater than 1, let p = log base x of (x + 1) and q = log base (x + 1) of (x + 2). Which of the following is correct?

Accepted Answer

Introduction / Context: This problem compares two logarithmic expressions whose bases and arguments both depend on x. It is designed to test a deeper understanding of how logarithms behave when both the base and the argument change, and how to reason about inequalities involving logs. Rather than plugging in random values, the goal is to understand the monotonic behaviour and show that one expression is always larger than the other for natural x greater than 1. Given Data / Assumptions: - x is a natural number and x is greater than 1. - p = log_x (x + 1). - q = log_{x + 1} (x + 2). - Bases x and x + 1 are greater than 1, so the logarithmic functions are increasing in their arguments. Concept / Approach: We convert both logarithms to a common base, for example base 10 or base e, using the change of base formula. Then we compare p and q numerically for general x. One useful idea is to express p and q in terms of ordinary logarithms of x, x + 1 and x + 2, and then analyse how the ratio of these logs behaves. Another more intuitive approach is to evaluate p and q for a few values of x (such as x = 2, 3, 4) and observe a consistent pattern, then reason why that pattern must hold for larger x as well. Step-by-Step Solution: Use the change of base formula: log_x (x + 1) = ln(x + 1) / ln x. Similarly, log_{x + 1} (x + 2) = ln(x + 2) / ln(x + 1). Therefore p = ln(x + 1) / ln x and q = ln(x + 2) / ln(x + 1). For x greater than 1, ln x, ln(x + 1) and ln(x + 2) are all positive. Consider x = 2 as a test value: p = ln 3 / ln 2, q = ln 4 / ln 3. Numerically, ln 3 / ln 2 is approximately 1.585, while ln 4 / ln 3 is approximately 1.262, so p > q at x = 2. For larger x, the ratio ln(x + 1) / ln x stays greater than 1 but slowly decreases, while ln(x + 2) / ln(x + 1) stays closer to 1 and is smaller than ln(x + 1) / ln x for all x greater than 1. This can be confirmed numerically or by more advanced analysis of the function f(x) = ln(x + 1) / ln x - ln(x + 2) / ln(x + 1), which remains positive for x greater than 1. Hence p is greater than q for all natural x greater than 1. Verification / Alternative check: Evaluating for several natural values gives a clear pattern. For x = 3, p = ln 4 / ln 3 which is about 1.262, and q = ln 5 / ln 4 which is about 1.161, so p > q again. For x = 4, p is about 1.161, while q is about 1.113. This consistent behaviour suggests that p always exceeds q. The trend arises because as the base grows, the relative increase from x to x + 1 is larger than the relative increase from x + 1 to x + 2 in logarithmic terms. Why Other Options Are Wrong: p < q: This contradicts both numerical checks and the analytical behaviour of the expressions. p = q: There is no natural x greater than 1 for which ln(x + 1) / ln x equals ln(x + 2) / ln(x + 1). cannot be determined: The relationship can be determined and is consistently p > q for all natural x greater than 1. Common Pitfalls: Some students may incorrectly think that since x, x + 1 and x + 2 are close, p and q should be equal or impossible to compare. Others may forget that when the base of a logarithm is greater than 1, the function is increasing in its argument, which affects the inequality direction. Another frequent mistake is to attempt to compare p and q by cross multiplying the logs without taking care of the domains and positivity, which can lead to sign errors. Using the change of base formula in a systematic way and checking a few concrete numerical examples helps build correct intuition. Final Answer: For all natural numbers x greater than 1, we have p > q.

Question 8

If log10 5 + log10 (5x + 1) = log10 (x + 5) + 1, find the value of x.

Accepted Answer

Introduction / Context: This question involves solving an equation that contains common logarithms with base 10. It checks understanding of standard log properties, especially the product rule and the interpretation of the constant 1 as log10 10. The unknown appears inside the arguments of the logarithms, so we must simplify the equation using log rules and then solve a resulting algebraic equation in x. This type of manipulation is very common in aptitude and competitive exams. Given Data / Assumptions: - The equation is log10 5 + log10 (5x + 1) = log10 (x + 5) + 1. - All logarithms are common logarithms (base 10). - The arguments of the logs must be positive, so 5x + 1 greater than 0 and x + 5 greater than 0, which implies x greater than -1/5 and x greater than -5. - Because we are interested in a real solution consistent with the answer options, we focus on positive x that satisfy these conditions. Concept / Approach: We use two main log identities: log10 A + log10 B = log10 (A * B) and 1 = log10 10. The presence of the constant 1 on the right side can be replaced with log10 10, so that every term in the equation becomes a log10 of some positive quantity. After that, we can combine logs on each side, compare their arguments, and obtain a simple linear equation in x. Solving that gives the required value. Step-by-Step Solution: Start with log10 5 + log10 (5x + 1) = log10 (x + 5) + 1. Rewrite 1 as log10 10, because log10 10 = 1. The equation becomes log10 5 + log10 (5x + 1) = log10 (x + 5) + log10 10. Use the product rule on the left: log10 5 + log10 (5x + 1) = log10 [5(5x + 1)]. Use the product rule on the right: log10 (x + 5) + log10 10 = log10 [10(x + 5)]. So we have log10 [5(5x + 1)] = log10 [10(x + 5)]. For positive arguments, equality of logs with the same base implies equality of the arguments. Therefore 5(5x + 1) = 10(x + 5). Expand: 25x + 5 = 10x + 50. Rearrange: 25x - 10x = 50 - 5, so 15x = 45. Hence x = 45 / 15 = 3. Verification / Alternative check: Substitute x = 3 back into the original equation. The left side is log10 5 + log10 (5(3) + 1) = log10 5 + log10 16. The right side is log10 (3 + 5) + 1 = log10 8 + 1. Numerically, log10 5 + log10 16 = log10 80, while log10 8 + 1 = log10 8 + log10 10 = log10 80. Both sides are equal, which confirms that x = 3 satisfies the equation. The domain conditions are also satisfied because 5x + 1 and x + 5 are positive when x = 3. Why Other Options Are Wrong: 1: If x = 1, then 5x + 1 = 6 and x + 5 = 6, giving log10 5 + log10 6 on the left and log10 6 + 1 on the right. These are not equal. 5: If x = 5, then 5x + 1 = 26 and x + 5 = 10, which leads to unequal expressions after applying the log rules. 10: If x = 10, then the expressions involve large and unequal arguments, which again do not satisfy the equation when checked numerically. Common Pitfalls: A typical mistake is forgetting to interpret the constant 1 as log10 10, causing students to attempt unnecessary operations like exponentiating both sides too early. Another error is combining logs incorrectly, for example treating log10 (x + 5) + 1 as log10 (x + 6) instead of log10 (10(x + 5)). Students also sometimes forget the domain restriction that arguments of logarithms must be positive, leading to extraneous solutions in more complex cases. Using the log product rule systematically and equating arguments only after combining logs on each side helps avoid these issues. Final Answer: The value of x that satisfies the equation is x = 3.

Question 9

What is the characteristic (integer part) of the common logarithm of the number 0.0000134?

Accepted Answer

Introduction / Context: This question tests the concept of the characteristic of a common logarithm. The characteristic is the integer part of log10 N, and it indicates the order of magnitude of the number in base 10. Understanding how to find the characteristic quickly, especially for numbers less than 1 and greater than 0, is important for working with logarithm tables and for solving digit and magnitude problems in aptitude exams. Given Data / Assumptions: - The number is 0.0000134. - We are dealing with the common logarithm, that is log10 of the number. - The characteristic is defined as the integer part of the logarithm, which will be negative for numbers between 0 and 1. Concept / Approach: For any positive number N, you can write it in scientific notation as N = m * 10^k, where 1 less than or equal to m less than 10 and k is an integer. Then log10 N = log10 m + k. The characteristic is the integer k when log10 m is between 0 and 1. For numbers greater than 1, k is non negative, and for numbers between 0 and 1, k is negative. The rule for a number between 0 and 1 is that if N lies between 10^(-n) and 10^(-n+1), then the characteristic is -n. Step-by-Step Solution: Write the number 0.0000134 in the form m * 10^k. Count the number of decimal places until the first non zero digit. The number 0.0000134 has 5 zeros after the decimal point before the first non zero digit 1. We can rewrite 0.0000134 = 1.34 * 10^-5. Now compute log10 (0.0000134) as log10 (1.34 * 10^-5). Use the product rule: log10 (1.34 * 10^-5) = log10 1.34 + log10 10^-5. We know log10 10^-5 = -5. The value log10 1.34 is between 0 and 1, so it does not change the integer part. Thus log10 (0.0000134) = (a small positive number) + (-5) = something between -5 and -4. Therefore the characteristic, which is the integer part of the logarithm, is -5. Verification / Alternative check: We can use the general rule for numbers between 0 and 1. A number of the form 0.0000...x, where there are n zeros after the decimal point before the first non zero digit, has a characteristic of -(n + 1). In this case, there are 4 zeros after the decimal point before the first non zero digit if we count carefully: 0.0000134 can be seen as 0.0000 134 with four zeros after the decimal point before 1, which gives characteristic -5. This shortcut is commonly used when working quickly with log tables. Why Other Options Are Wrong: 5: This would be the characteristic of a large number near 10^5, not a small number less than 1. 6: This would correspond to an even larger number of the order of 10^6, which is completely inconsistent with 0.0000134. -6: This would indicate a much smaller number of order 10^-6 or less, whereas 0.0000134 is between 10^-5 and 10^-4, not as small as 10^-6. Common Pitfalls: A frequent mistake is to miscount the number of zeros after the decimal point or to think that the number of zeros directly gives the characteristic without adding or subtracting one. Another error is to confuse the characteristic with the mantissa. The mantissa is the fractional part of the logarithm and is always non negative when using standard logarithm tables. Here we only need the integer part, which is negative for numbers between 0 and 1. Carefully writing the number in scientific notation helps avoid these errors. Final Answer: The characteristic of log10 (0.0000134) is -5.

Question 10

Find the value of log10 (tan 10°) + log10 (tan 20°) + log10 (tan 30°) + log10 (tan 40°) + log10 (tan 50°) + log10 (tan 60°) + log10 (tan 70°) + log10 (tan 80°).

Accepted Answer

Introduction / Context: This question combines trigonometry and logarithms. It asks for the value of a sum of logarithms of tangent values at various angles. Instead of computing each tangent and each logarithm separately, which would be tedious and inaccurate, we should use trigonometric identities and log properties to simplify the entire expression. The key idea is the complementary angle identity for tangent, which leads to products equal to 1 and hence logarithms equal to 0. Given Data / Assumptions: - We must evaluate S = log10 (tan 10°) + log10 (tan 20°) + log10 (tan 30°) + log10 (tan 40°) + log10 (tan 50°) + log10 (tan 60°) + log10 (tan 70°) + log10 (tan 80°). - All logarithms are common logarithms with base 10. - The tangent function is defined for these angles in degrees, and none of these angles is an odd multiple of 90°, so tan values are finite. Concept / Approach: The main trigonometric identity we use is tan θ * tan (90° - θ) = 1 for angles where both tangent values are defined. Taking logarithms of both sides gives log10 (tan θ) + log10 (tan (90° - θ)) = log10 1 = 0. This means that each pair of complementary angles contributes zero to the total sum of logs. The given angles naturally form such complementary pairs: 10° with 80°, 20° with 70°, 30° with 60°, and 40° with 50°. Step-by-Step Solution: Observe the complementary angle pairs: (10°, 80°), (20°, 70°), (30°, 60°), (40°, 50°). Use the identity tan θ * tan (90° - θ) = 1 for each pair. For θ = 10°, tan 10° * tan 80° = 1. Taking common logs: log10 (tan 10°) + log10 (tan 80°) = log10 1 = 0. Similarly, log10 (tan 20°) + log10 (tan 70°) = 0. Also, log10 (tan 30°) + log10 (tan 60°) = 0. And log10 (tan 40°) + log10 (tan 50°) = 0. Now group the original sum S into these four pairs. S = [log10 (tan 10°) + log10 (tan 80°)] + [log10 (tan 20°) + log10 (tan 70°)] + [log10 (tan 30°) + log10 (tan 60°)] + [log10 (tan 40°) + log10 (tan 50°)]. Each bracket equals 0, so S = 0 + 0 + 0 + 0 = 0. Verification / Alternative check: As a numerical check, we can approximate a few values. For example, tan 30° is 1 / sqrt(3) and tan 60° is sqrt(3). Their product is 1, so log10 (tan 30°) + log10 (tan 60°) = log10 1 = 0, matching the theoretical identity. Likewise, tan 45° is 1, and although 45° does not appear here, it serves as a reminder that tangent and its complementary values often form simple pairs. Using a calculator for a few pairs confirms that each pair of logs sums to zero within rounding error. Why Other Options Are Wrong: -1: This would require the overall product of all tan values to be 10^-1, which contradicts the pair wise product being 1 for each complementary pair. 1/2 or 1: These non zero values would imply that the product of all tan values is a power of 10 different from 1, again contradicting the identity tan θ * tan (90° - θ) = 1 for each pair. Common Pitfalls: Students may overlook the complementary angle pattern and attempt to compute each term individually, which is slow and error prone. Another common mistake is to apply the sine or cosine complementary identities instead of the tangent identity, or to incorrectly think that tan (90° - θ) equals cot θ without using the full product identity. Forgetting that log10 1 equals 0 can also lead to confusion. Recognizing symmetry in angle sets and pairing complementary angles is the key to these problems. Final Answer: The value of the given sum of logarithms is 0.

Question 11

Evaluate the expression (1 / 3) log10 125 - 2 log10 4 + log10 32.

Accepted Answer

Introduction / Context: This question focuses on evaluating an expression involving common logarithms by applying standard log properties such as the power rule and the product and quotient rules. It helps students practise simplifying log expressions without resorting to calculators. These skills are very important in aptitude tests and competitive exams that emphasise algebraic manipulation. Given Data / Assumptions: - The expression is E = (1 / 3) log10 125 - 2 log10 4 + log10 32. - All logarithms are common logarithms (base 10). - The numbers 125, 4 and 32 are positive, so the logs are defined. Concept / Approach: We use three basic properties of logs: 1. Power rule: k log10 A = log10 (A^k). 2. Sum rule: log10 A + log10 B = log10 (A * B). 3. Difference rule: log10 A - log10 B = log10 (A / B). By rewriting each term using the power rule, we can combine all terms into a single log10 of a simple number. Then we evaluate that number directly to find the final value. Step-by-Step Solution: Start with E = (1 / 3) log10 125 - 2 log10 4 + log10 32. Apply the power rule to the first term: (1 / 3) log10 125 = log10 (125^(1 / 3)). Since 125 = 5^3, its cube root 125^(1 / 3) is 5, so (1 / 3) log10 125 = log10 5. Apply the power rule to the second term: -2 log10 4 = log10 (4^-2). Compute 4^-2 = 1 / 4^2 = 1 / 16. So the expression becomes E = log10 5 + log10 32 + log10 (4^-2). Use the sum rule to combine: E = log10 [5 * 32 * 4^-2]. Compute 4^2 = 16, so 4^-2 = 1 / 16. Thus E = log10 [5 * 32 / 16]. Calculate the product and quotient: 32 / 16 = 2, so 5 * 2 = 10. Hence E = log10 10. We know log10 10 = 1. Verification / Alternative check: We can also express everything in terms of powers of 2 and 5. Note that 125 = 5^3, 4 = 2^2, and 32 = 2^5. Rewriting E as (1 / 3) log10 (5^3) - 2 log10 (2^2) + log10 (2^5) and applying the power rule directly gives log10 5 - 4 log10 2 + 5 log10 2 = log10 5 + log10 2 = log10 10 = 1. This matches our previous result, confirming the answer. Why Other Options Are Wrong: 0: This would require the combined product inside the final logarithm to be 1, which it is not. 2: This would correspond to log10 of 100, but we found that the effective argument is 10, not 100. 4/5: This is not consistent with any simple power of 10 and does not match the simplified log expression. Common Pitfalls: Students sometimes forget to apply the power rule correctly and treat (1 / 3) log10 125 as log10 (125 / 3), which is wrong. Another common error is mishandling negative exponents and incorrectly simplifying 4^-2. Misapplication of the sum and difference rules can also lead to combining terms in the wrong way. Keeping the three basic log rules clear and working step by step avoids these problems. Final Answer: The value of the expression is 1.

Question 12

If log base 3 of 30 is equal to 1 divided by a and log base 5 of 30 is equal to 1 divided by b, that is log_3 30 = 1 / a and log_5 30 = 1 / b, then find the value of 3 log base 30 of 2 in terms of a and b.

Accepted Answer

Introduction / Context: This algebraic logarithm question asks you to express 3 log base 30 of 2 in terms of two parameters a and b that are defined via log_3 30 and log_5 30. It tests the ability to use the change of base formula, to manipulate logarithmic expressions symbolically, and to express one log in terms of others using factorisation of the number 30. Such symbolic log problems are common in algebra and aptitude practice sets. Given Data / Assumptions: - log_3 30 = 1 / a. - log_5 30 = 1 / b. - All numbers involved (2, 3, 5, 30) are positive and different from 1, so all required logs are defined. - We must find the expression for 3 log_30 2 in terms of a and b. Concept / Approach: We will rewrite each given relation in terms of a single common base, for example base 10 or base e, using the change of base formula log_p q = ln q / ln p. Then we can express ln 3 and ln 5 in terms of ln 30, a and b. Because 30 factorises as 2 * 3 * 5, we can find ln 2 in terms of ln 30, ln 3 and ln 5. Finally we use log_30 2 = ln 2 / ln 30 and multiply by 3 to obtain the desired expression. Step-by-Step Solution: From log_3 30 = 1 / a, apply change of base: log_3 30 = ln 30 / ln 3. So ln 30 / ln 3 = 1 / a, which implies a = ln 3 / ln 30. Thus ln 3 = a ln 30. From log_5 30 = 1 / b, apply change of base: log_5 30 = ln 30 / ln 5. So ln 30 / ln 5 = 1 / b, which implies b = ln 5 / ln 30. Thus ln 5 = b ln 30. Since 30 = 2 * 3 * 5, take logs: ln 30 = ln 2 + ln 3 + ln 5. Substitute ln 3 and ln 5 in terms of ln 30: ln 30 = ln 2 + a ln 30 + b ln 30. Rearrange to solve for ln 2: ln 2 = ln 30 - a ln 30 - b ln 30 = (1 - a - b) ln 30. Now log_30 2 = ln 2 / ln 30 = (1 - a - b) ln 30 / ln 30 = 1 - a - b. So 3 log_30 2 = 3 (1 - a - b). Verification / Alternative check: We can check consistency by choosing a specific numerical base to compute approximate values. Let ln denote the natural log. Evaluate log_3 30 numerically, find a, and then compute 3 log_30 2 directly. Comparing it with 3 (1 - a - b) after also computing b from log_5 30 shows that the two sides are equal within rounding error. This provides a numerical verification of the algebraic manipulation. Why Other Options Are Wrong: 3(1 + a + b): This would arise if ln 2 were incorrectly taken as (1 + a + b) ln 30, which is inconsistent with the factorisation equation ln 30 = ln 2 + ln 3 + ln 5. 2(1 - a - b): This has the correct combination 1 - a - b but an incorrect coefficient 2 instead of 3. The factor of 3 comes directly from 3 log_30 2. 3(1 + a - b): This corresponds to an incorrect expression for ln 2 where ln 5 is subtracted rather than added. That contradicts the identity ln 30 = ln 2 + ln 3 + ln 5. Common Pitfalls: A common error is to confuse log_3 30 with log_30 3 and hence invert a and 1 / a incorrectly. Another mistake is to ignore the factor 3 in 3 log_30 2 until the end and then forget to multiply the final result. Students also sometimes mishandle the factorisation 30 = 2 * 3 * 5 and write ln 30 = ln 3 + ln 5 - ln 2, which is wrong. Being systematic with change of base and carefully using the factorisation identity prevents these errors. Final Answer: The value of 3 log base 30 of 2 is 3(1 - a - b).

Question 13

Which of the following statements about common logarithms (base 10) is not correct?

Accepted Answer

Introduction / Context: This conceptual question checks whether students understand the basic properties of common logarithms and know when addition inside a log can or cannot be simplified. Many errors in log problems come from misusing sum and product rules, so identifying the incorrect statement is an important skill. All logs here are common logs, that is logarithms to base 10. Given Data / Assumptions: We are given four statements: - A: log 10 = 1. - B: log (2 + 3) = log (2 x 3). - C: log 1 = 0. - D: log (1 + 2 + 3) = log 1 + log 2 + log 3. All logs are base 10, and we must identify which statement is not correct. Concept / Approach: The key log properties are: 1. log 10 = 1 because 10^1 = 10. 2. log 1 = 0 because 10^0 = 1. 3. log (A B) = log A + log B, but log (A + B) is not equal to log A + log B in general. We evaluate each option using these rules. We check whether each side represents the same quantity. If any equality incorrectly treats a sum as if it were a product, that statement is invalid. Step-by-Step Solution: Check option A: log 10 = 1. Since 10^1 = 10, this is correct. Check option C: log 1 = 0. Since 10^0 = 1, this is also correct. Check option B: log (2 + 3) = log (2 x 3). Left side is log 5 and right side is log 6. Because 5 is not equal to 6, log 5 is not equal to log 6, so option B is incorrect. Check option D: log (1 + 2 + 3) = log 1 + log 2 + log 3. Left side is log 6. Right side is log 1 + log 2 + log 3 = 0 + log 2 + log 3 = log (2 x 3) = log 6, using the product rule. Therefore option D is correct because both sides equal log 6. Verification / Alternative check: Numerically, we can approximate the logs. Using base 10, log 5 is about 0.6990, while log 6 is about 0.7782. Clearly these are not equal, which confirms that log (2 + 3) is not equal to log (2 x 3). For statement D, log 2 is about 0.3010 and log 3 is about 0.4771, so log 2 + log 3 is about 0.7781, which matches log 6 up to rounding. This confirms that option D is correct, whereas option B is not. Why Other Options Are Wrong: The question asks for the statement that is not correct, so only option B is wrong. - Option A is consistent with the fundamental definition of the common logarithm of 10. - Option C is consistent with the fundamental definition of the common logarithm of 1. - Option D correctly uses the product rule, because 1 x 2 x 3 equals 6, and the log of the product equals the sum of the logs. Common Pitfalls: A very common mistake is to treat log (A + B) as log A + log B, which is never generally true. This leads to errors in many algebraic and aptitude problems. Another pitfall is not recognising that log (1 + 2 + 3) is log 6, and that log 1 + log 2 + log 3 simplifies to log 6 via the product rule. Students may also assume that any equality involving logs is correct if the numbers look small and simple. It is crucial to remember exactly when the log of a product equals the sum of logs and that there is no simple rule for the log of a sum. Final Answer: The statement that is not correct is log (2 + 3) = log (2 x 3).

Question 14

Find the value of log base 5 of 5 multiplied by log base 4 of 9 multiplied by log base 3 of 2, that is log_5 5 × log_4 9 × log_3 2.

Accepted Answer

Introduction / Context: This question tests the understanding of how logarithms with different bases can simplify when they are multiplied together. The expression involves three logs with bases 5, 4 and 3 respectively, and we must evaluate their product. Recognising which parts simplify to 1 and how to use the change of base formula is the key to solving this quickly in an exam. Given Data / Assumptions: - The expression to evaluate is E = log_5 5 × log_4 9 × log_3 2. - The bases 5, 4 and 3 are positive numbers not equal to 1, so all logs are defined. - We are looking for a numerical value, not a symbolic expression. Concept / Approach: First, note that log_5 5 is equal to 1 because any log of the same base and argument equals 1. Next, we relate log_4 9 and log_3 2. By writing them in terms of a common base using the change of base formula, we can see that log_4 9 and log_3 2 are reciprocals. The product of a number and its reciprocal is 1, and since the first factor is already 1, the entire expression simplifies to 1. Step-by-Step Solution: Start with E = log_5 5 × log_4 9 × log_3 2. Observe that log_5 5 = 1 because 5^1 = 5. So E = 1 × log_4 9 × log_3 2 = log_4 9 × log_3 2. Use change of base for log_4 9: log_4 9 = ln 9 / ln 4. Write 9 and 4 as powers: 9 = 3^2 and 4 = 2^2. Then ln 9 = 2 ln 3 and ln 4 = 2 ln 2, so log_4 9 = (2 ln 3) / (2 ln 2) = ln 3 / ln 2. Use change of base for log_3 2: log_3 2 = ln 2 / ln 3. Now multiply: log_4 9 × log_3 2 = (ln 3 / ln 2) × (ln 2 / ln 3). The ln 3 and ln 2 terms cancel, leaving 1. Therefore E = 1. Verification / Alternative check: We can numerically approximate the values to confirm. log_5 5 is exactly 1. log_4 9 can be approximated as ln 9 / ln 4, and log_3 2 as ln 2 / ln 3. Multiplying these two approximations gives a result extremely close to 1.0, confirming the algebraic simplification. This shows that both the symbolic reasoning and numerical approximation agree. Why Other Options Are Wrong: 3/2: This would require log_4 9 × log_3 2 to be 1.5, which contradicts the reciprocal relationship between log_4 9 and log_3 2. 2: This would occur only if log_4 9 and log_3 2 were both equal to the same number greater than 1, which is not true. 5: There is no factor of 5 that remains after simplification, so this is unrelated to the correct product. Common Pitfalls: Students sometimes forget that log_a a is 1 and treat it as an unknown. Another common mistake is to misapply the change of base formula, for example writing log_4 9 as ln 4 / ln 9. This reverses the ratio and destroys the reciprocal relationship with log_3 2. Some may also try to approximate each log numerically without simplification, which is slower and prone to rounding error. Systematically using change of base with the correct numerator and denominator avoids these issues. Final Answer: The value of log_5 5 × log_4 9 × log_3 2 is 1.

Question 15

Given log10 2 = 0.301 and log10 3 = 0.477, how many digits are there in the decimal representation of 108 raised to the power 10?

Accepted Answer

Introduction / Context: This question asks for the number of decimal digits in a large power, 108^10, using approximate values of log10 2 and log10 3. It is a classic application of logarithms to digit counting and is commonly seen in competitive exams. The idea is to avoid direct computation of the power by using the relationship between the common logarithm and the number of digits of a positive integer. Given Data / Assumptions: - log10 2 = 0.301 (approximate value). - log10 3 = 0.477 (approximate value). - We must find the number of digits in 108^10. - All logarithms are common logarithms in base 10. Concept / Approach: If N is a positive integer, the number of digits in N in base 10 is given by digits = floor(log10 N) + 1. Here N is 108^10. To compute log10 N, we first express 108 in terms of its prime factors 2 and 3, then use the power rule of logarithms. Since 108 = 2^2 * 3^3, we can find log10 108 in terms of log10 2 and log10 3. Then log10 108^10 = 10 * log10 108. Finally, applying the digit formula gives the required number of digits. Step-by-Step Solution: First factor 108: 108 = 2^2 * 3^3. Compute log10 108 using the product rule and the power rule. log10 108 = log10 (2^2 * 3^3). Apply the product rule: log10 (2^2 * 3^3) = log10 (2^2) + log10 (3^3). Use the power rule: log10 (2^2) = 2 log10 2 and log10 (3^3) = 3 log10 3. Substitute the given values: log10 108 = 2(0.301) + 3(0.477). Compute: 2(0.301) = 0.602 and 3(0.477) = 1.431. So log10 108 is approximately 0.602 + 1.431 = 2.033. Now compute log10 (108^10) = 10 * log10 108. Thus log10 (108^10) is approximately 10 * 2.033 = 20.33. The number of digits in 108^10 is floor(20.33) + 1 = 20 + 1 = 21. Verification / Alternative check: We know that 10^20 has 21 digits and 10^21 has 22 digits. The value 108^10 has log10 value about 20.33, which lies between 20 and 21. This implies that 10^20 < 108^10 < 10^21, so 108^10 must have more digits than 10^20 but fewer digits than 10^21. Therefore it has exactly 21 digits. This confirms the result obtained from the floor and plus one formula and shows that small variations in the decimal approximations of log10 2 and log10 3 will not change the digit count. Why Other Options Are Wrong: 19: This would require log10 (108^10) to be between 18 and 19, which is far from the computed value of about 20.33. 20: This would require log10 (108^10) to be between 19 and 20, which again does not match 20.33. 22: This would require log10 (108^10) to be at least 21, which is larger than the computed 20.33, so it is impossible. Common Pitfalls: Students sometimes round intermediate values too aggressively, for example rounding log10 108 to 2 instead of 2.033, which may still give the correct digit count but can be risky in closer cases. Another error is to forget the plus one in the digit formula and answer floor(log10 N) instead of floor(log10 N) + 1. Some also misfactor 108, for example writing 108 = 2^3 * 3^2, which would lead to an incorrect log10 108 and hence an incorrect digit count. Carefully factoring 108 and applying the log rules step by step prevents these mistakes. Final Answer: The number of digits in 108^10 is 21.

Question 16

Find the value of log base 2 of (log base 5 of 625), that is log_2 (log_5 625).

Accepted Answer

Introduction / Context: This question asks for the value of a nested logarithm, where one logarithm appears inside another. It tests the understanding of evaluating simple logs with integer powers and then using that result as the argument for another log with a different base. Recognising perfect powers of the base and using the definition of logarithms makes this problem straightforward. Given Data / Assumptions: - We must evaluate log_2 (log_5 625). - All logs are real and their arguments and bases are positive and bases are not equal to 1. - 625 is a positive number and clearly related to base 5. Concept / Approach: We evaluate the inner logarithm log_5 625 first, using the definition of a logarithm: log_b a is the exponent to which b must be raised to obtain a. Because 625 is a power of 5, this inner log is an integer. Then we take the outer logarithm base 2 of that integer. Again, since the resulting number is a power of 2, the outer log simplifies to an integer as well. Step-by-Step Solution: Start with the inner log: log_5 625. Express 625 as a power of 5. We have 625 = 5^4. By definition, log_5 (5^4) = 4. So log_5 625 = 4. Now the original expression becomes log_2 (log_5 625) = log_2 4. Express 4 as a power of 2. We have 4 = 2^2. By definition, log_2 (2^2) = 2. Therefore log_2 (log_5 625) = 2. Verification / Alternative check: We can check numerically using basic knowledge of exponentials. Since 5^4 = 625, the inner log is exactly 4. The outer log asks for the exponent that produces 4 when the base is 2. Because 2^2 = 4, the outer log is exactly 2. No approximations are needed, and both steps use the direct definition of the logarithm function. Why Other Options Are Wrong: 5: This would suggest that 2^5 equals 4, which is false because 2^5 equals 32. 10: This would suggest that 2^10 equals 4, which is false because 2^10 equals 1024. 15: This would be even larger and has no relation to the small numbers 4 and 2 appearing in the simplification. Common Pitfalls: Some students try to apply change of base formulas immediately and involve unnecessary decimal approximations. Others misidentify 625 as 5^3 instead of 5^4 or confuse log_5 625 with log_625 5. Another error is to compute log_2 5 or log_5 2 in the wrong place, which is not needed here. Focusing on expressing 625 and 4 as simple powers of their respective bases makes the problem quick and exact. Final Answer: The value of log_2 (log_5 625) is 2.

Question 17

If a = b^x, b = c^y and c = a^z for positive real numbers a, b and c not equal to 1, what is the value of the product x y z?

Accepted Answer

Introduction / Context: This question involves exponential relationships between three positive real numbers a, b and c. Each number is expressed as a power of another, and we are asked to find the value of the product x y z. It tests the ability to work with logarithms and exponents in a system of equations and to recognise how cyclic relationships can force a simple condition on the exponents. Given Data / Assumptions: - a = b^x. - b = c^y. - c = a^z. - The numbers a, b and c are positive and not equal to 1. - We must find the numerical value of x y z that always holds for such a system. Concept / Approach: We use logarithms to convert the exponential equations into linear equations in terms of log a, log b and log c. This makes the relationships easier to handle algebraically. After expressing log a, log b and log c in terms of one another and the exponents x, y and z, we combine the three equations in a way that isolates the product x y z. The crucial step is multiplying the equations so that the logs cancel and we can solve for x y z as a constant. Step-by-Step Solution: Start with the three equations: a = b^x, b = c^y and c = a^z. Take natural logs (or any consistent log base) of all three equations. From a = b^x, we get ln a = x ln b. From b = c^y, we get ln b = y ln c. From c = a^z, we get ln c = z ln a. Now substitute ln b from the second equation into the first: ln a = x (y ln c) = x y ln c. Then substitute ln c from the third equation: ln a = x y (z ln a) = x y z ln a. So ln a = x y z ln a. Since a is positive and not equal to 1, ln a is not zero. Divide both sides by ln a to obtain 1 = x y z. Therefore x y z = 1. Verification / Alternative check: To verify, we can construct a simple example. Let a, b and c all be equal to 2. Then a = b^x becomes 2 = 2^x so x = 1. Similarly b = c^y becomes 2 = 2^y so y = 1, and c = a^z becomes 2 = 2^z so z = 1. Then x y z = 1 * 1 * 1 = 1, matching the derived condition. If we choose any other positive a, b, c not equal to 1 that satisfy the given cyclic relations, the same logarithmic argument will show that x y z must always be 1. Why Other Options Are Wrong: -1: This would imply that the product of the exponents is negative, which contradicts the derived equation ln a = x y z ln a for non zero ln a. 0: This would require at least one of x, y or z to be zero, which would break one of the original exponential relationships unless one of a, b or c were equal to 1, which is not allowed. abc: This is a product of the bases, not of the exponents, and has no support from the logarithmic analysis of the system. Common Pitfalls: A common mistake is to attempt to substitute the expressions directly in their exponential form without using logarithms, which can become messy and confusing. Another error is to cancel terms incorrectly when manipulating exponents, especially when raising one equation to another power. Some students also forget the condition that a, b and c are not equal to 1, which ensures that ln a, ln b and ln c are non zero and division is valid. Using logarithms systematically and recalling that logs convert exponents into multipliers leads directly to the simple result x y z = 1. Final Answer: The value of the product x y z is 1.

Question 18

If a, b and c are respectively the p-th, q-th and r-th terms of a geometric progression (G.P.), then what is the value of the expression (q − r) log a + (r − p) log b + (p − q) log c ?

Accepted Answer

Introduction / Context: This question tests an important identity involving logarithms and the general term of a geometric progression. The terms a, b and c are taken from the same geometric progression at positions p, q and r respectively, and we are asked to simplify a weighted sum of logarithms of these terms. Such identities are common in algebra and aptitude exams, because they check understanding of both sequences and the properties of logarithms. Given Data / Assumptions: a is the p-th term of a geometric progression.b is the q-th term of the same geometric progression.c is the r-th term of the same geometric progression.We consider the expression S = (q − r) log a + (r − p) log b + (p − q) log c.Logarithm base is assumed to be the same throughout, so properties of logarithms apply. Concept / Approach: In a geometric progression, the n-th term can be written as Tn = A * R^(n − 1), where A is the first term and R is the common ratio. Taking logarithms converts multiplication and powers into addition and products, which makes algebraic manipulation easier. By expressing log a, log b and log c in terms of log A and log R, and then substituting into S, many terms cancel out and the expression simplifies significantly. Step-by-Step Solution: Step 1: Let the geometric progression have first term A and common ratio R.Step 2: Then a = A * R^(p − 1), b = A * R^(q − 1), c = A * R^(r − 1).Step 3: Take logarithms: log a = log A + (p − 1) log R.Step 4: Similarly, log b = log A + (q − 1) log R and log c = log A + (r − 1) log R.Step 5: Substitute these into S = (q − r) log a + (r − p) log b + (p − q) log c.Step 6: Collect coefficients of log A: (q − r + r − p + p − q) log A = 0 * log A = 0.Step 7: Collect coefficients of log R and simplify; they also sum to zero after algebraic cancellation.Step 8: Therefore S reduces to 0 in every case, regardless of A, R, p, q and r, provided the progression is valid. Verification / Alternative check: To verify, take a simple geometric progression such as 2, 4, 8, 16, 32 and pick p = 1, q = 2, r = 3, so a = 2, b = 4 and c = 8. Compute (q − r) log a + (r − p) log b + (p − q) log c numerically. After substituting the values and simplifying, the sum again comes out to zero. Trying a different progression, for example 3, 6, 12, 24, leads to the same result, which confirms that the simplification is independent of the particular geometric sequence chosen. Why Other Options Are Wrong: The option 1 would mean the expression always evaluates to a constant nonzero value, which contradicts direct substitution in examples. The option −1 is also inconsistent with test values. The option pqr depends on the product of indices and would vary when p, q and r change, but the simplified expression is independent of particular index values. Only the value 0 matches both the algebraic identity and numerical checks. Common Pitfalls: A common mistake is to confuse arithmetic and geometric progressions and write terms in the wrong form. Another error is to forget that logarithms are taken with the same base and thus properties like log(X * Y) = log X + log Y apply consistently. Some learners also mishandle algebra when collecting coefficients of log A and log R and fail to notice that these coefficients sum to zero. Careful symbolic manipulation avoids these mistakes. Final Answer: The simplified value of the expression is 0, so the correct option is the one that states zero.

Question 19

The value of the logarithm log base 343 of 7, that is log_343 7, is equal to which of the following?

Accepted Answer

Introduction / Context: This question checks understanding of logarithms with unusual bases and arguments. Instead of simple base 10 or base e, the base is 343 and the argument is 7. Problems like this appear in aptitude tests to see whether the candidate can recognize powers and rewrite a logarithm in terms of exponents without relying on a calculator. Given Data / Assumptions: We are asked to find log base 343 of 7.The expression is written as log_343 7.The usual change of base and exponent properties of logarithms are available.We work with real positive numbers only. Concept / Approach: The key idea is to express both the base and the argument as powers of the same prime number. The number 343 can be written as 7^3. The number 7 is 7^1. When both base and argument share the same prime factor, the logarithm log_(a^k) (a^m) simplifies neatly to m / k. This comes from the definition of logarithm: log_b x is the exponent to which b must be raised to produce x. Step-by-Step Solution: Step 1: Rewrite the base 343 as a power of 7: 343 = 7^3.Step 2: Rewrite the argument 7 as 7^1.Step 3: Now consider log_343 7 as log_(7^3) (7^1).Step 4: By the definition of logarithm, we seek the exponent y such that (7^3)^y = 7^1.Step 5: Simplify the left side: (7^3)^y = 7^(3y).Step 6: Equate exponents since the bases are equal: 3y = 1.Step 7: Solve for y: y = 1 / 3.Step 8: Therefore log_343 7 = 1 / 3. Verification / Alternative check: We can confirm the result by using the change of base formula. Write log_343 7 = log 7 / log 343, where log denotes any consistent base, for example base 10. Then express 343 as 7^3. Hence log 343 = log (7^3) = 3 log 7. So log_343 7 = log 7 / (3 log 7) = 1 / 3. This matches the earlier exponent based reasoning and confirms that the correct value is one third. Why Other Options Are Wrong: The option 3 would correspond to log_7 343, not log_343 7. The option −3 and the option −1 / 3 are negative, but both base and argument are greater than 1, so the logarithm must be positive. Therefore, all negative values are impossible. Only 1 / 3 fits the algebraic derivation and the general behavior of logarithms. Common Pitfalls: A common error is to invert the roles of base and argument and mistakenly compute log_7 343. Another mistake is to overlook that 343 is a power of 7 and attempt to use approximate decimal values, which complicates the problem unnecessarily. Recognizing power relationships and using exponent rules keeps the solution short and accurate. Final Answer: The value of log base 343 of 7 is 1/3, so the correct option is 1/3.

Question 20

The greatest possible natural number n that satisfies the inequality 9^n < 10^8, given that log 3 = 0.4771 and n ∈ N, is equal to:

Accepted Answer

Introduction / Context: This question involves comparing an exponential expression with a power of 10 and finding the largest natural number n that satisfies a given inequality. Logarithms are provided as a tool for handling powers, particularly because the base of the exponent is 9 and the comparison base is 10. Problems of this form are common in aptitude tests to check fluency with logarithmic transformations and inequalities. Given Data / Assumptions: We need the largest natural number n such that 9^n < 10^8.We are given log 3 = 0.4771 (base 10).The variable n is a natural number, so n is a positive integer.We use base 10 logarithms for convenience. Concept / Approach: The inequality compares 9^n with 10^8. Taking logarithms on both sides turns powers into products, which makes it easier to isolate n. Since 9 = 3^2, we can rewrite 9^n as 3^(2n). Using log rules, log (a^k) = k log a. The inequality then becomes linear in n, and we can solve for n using the given logarithm of 3. Finally, as n must be an integer, we choose the largest integer that still satisfies the inequality. Step-by-Step Solution: Step 1: Rewrite 9^n using base 3: 9 = 3^2, so 9^n = (3^2)^n = 3^(2n).Step 2: The inequality becomes 3^(2n) < 10^8.Step 3: Take base 10 logarithms of both sides: log(3^(2n)) < log(10^8).Step 4: Apply log power rule: 2n log 3 < 8 log 10.Step 5: Since log 10 = 1 in base 10, this simplifies to 2n log 3 < 8.Step 6: Substitute the given value log 3 = 0.4771: 2n * 0.4771 < 8.Step 7: Compute 2 * 0.4771 = 0.9542, so 0.9542 n < 8.Step 8: Solve for n: n < 8 / 0.9542, which is approximately n < 8.38.Step 9: Since n must be a natural number, the greatest integer less than 8.38 is 8. Verification / Alternative check: As a check, evaluate 9^8 and 9^9. The value 9^8 is 43046721, which is less than 10^8 (that is, 100000000). The value 9^9 is 387420489, which is greater than 10^8. Therefore 9^8 < 10^8 but 9^9 ≥ 10^8. This confirms that n = 8 is the greatest natural number that satisfies the inequality. Why Other Options Are Wrong: The option 7 is too small; it satisfies the inequality but is not the greatest possible value. The options 9 and 10 suggest values that produce powers of 9 that are larger than 10^8, so they violate the inequality. Only n = 8 satisfies the inequality and is the largest such natural number. Common Pitfalls: One frequent mistake is to ignore the requirement that n be a natural number and to give a decimal answer like 8.38. Another error is to forget that when taking logarithms, the direction of the inequality remains unchanged because log is an increasing function for positive numbers. Some candidates also forget that log 10 is equal to 1 in base 10, which can lead to incorrect simplification. Final Answer: The greatest possible natural number n that satisfies 9^n < 10^8 is 8.

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