Question 1

If the linear equation 5x − 17 = −x + 7 holds, then what is the numerical value of x?

Accepted Answer

Introduction / Context:This question is a straightforward linear equation in one variable. It checks basic algebra skills, such as combining like terms and isolating the variable x. Such questions are common in aptitude and entrance tests because they quickly reveal whether a candidate is comfortable manipulating simple equations. Given Data / Assumptions:We are given the equation 5x − 17 = −x + 7.The unknown variable is x.All operations are over real numbers. Concept / Approach:The standard procedure for solving a linear equation is to bring all terms containing x to one side of the equation and constants to the other side. After that, we simplify and divide by the coefficient of x to find its value. Careful handling of negative signs is important to avoid arithmetic errors. Step-by-Step Solution:Step 1: Start from 5x − 17 = −x + 7.Step 2: Add x to both sides to collect x terms on the left: 5x + x − 17 = 7.Step 3: Simplify the x terms: 6x − 17 = 7.Step 4: Add 17 to both sides to move the constant: 6x = 7 + 17.Step 5: Compute the sum on the right: 6x = 24.Step 6: Divide both sides by 6: x = 24 / 6.Step 7: Simplify the fraction: x = 4. Verification / Alternative check:Substitute x = 4 back into the original equation. The left hand side becomes 5 * 4 − 17 = 20 − 17 = 3. The right hand side becomes −4 + 7 = 3. Since both sides are equal, the equation is satisfied for x = 4. This confirms that the solution is correct. Why Other Options Are Wrong:If x = 5, then 5x − 17 = 25 − 17 = 8 while −x + 7 = −5 + 7 = 2, which do not match. If x = 7, then 5x − 17 = 35 − 17 = 18 while −x + 7 = −7 + 7 = 0, again not equal. If x = 9, then 5x − 17 = 45 − 17 = 28 while −x + 7 = −9 + 7 = −2, which also does not satisfy the equation. Therefore only x = 4 works. Common Pitfalls:Learners sometimes subtract x from 5x incorrectly or mis-handle the constant term when moving it from one side to the other. Forgetting to perform the same operation on both sides of the equation is another common error. Double checking each step and performing a quick substitution at the end helps avoid these mistakes. Final Answer:The value of x that satisfies 5x − 17 = −x + 7 is 4.

Question 2

If A = log base 32 of 1875 and B = log base 243 of 2187, then which of the following correctly describes the relationship between A and B?

Accepted Answer

Introduction / Context:This question compares two logarithmic expressions with different bases and arguments. The aim is not to compute exact decimal values but to determine which of A or B is larger. Such comparisons are common in aptitude tests because they require factorization skills and a good understanding of how logarithms behave with respect to their bases and arguments. Given Data / Assumptions:A = log base 32 of 1875, written as log_32 1875.B = log base 243 of 2187, written as log_243 2187.All numbers involved are positive, and bases are not equal to 1. Concept / Approach:The key idea is to express all numbers in terms of their prime factorization, preferably using the same prime. For expression B, both 243 and 2187 are powers of 3, which allows an exact simplification. For expression A, while 32 is a power of 2, 1875 can be factorized to show its structure, and we can approximate its value or compare using change of base and inequalities. Once both expressions are in a manageable form, we can decide whether A is greater than, equal to or less than B. Step-by-Step Solution:Step 1: Factorize 243 and 2187. We have 243 = 3^5 and 2187 = 3^7.Step 2: Write B in terms of base 3: B = log_(3^5) (3^7).Step 3: By the exponent rule for logarithms, log_(a^k) (a^m) = m / k, so B = 7 / 5 = 1.4.Step 4: For A, note that 32 = 2^5 and factorize 1875 as 3 * 5^4.Step 5: Use the change of base formula: A = log 1875 / log 32 (any consistent log base).Step 6: Numerically, 1875 is much larger than 32, so the logarithm to base 32 of 1875 should be greater than 1. Indeed, calculation gives approximately A ≈ 2.17.Step 7: Compare the approximate values: A ≈ 2.17 while B = 1.4.Step 8: Therefore A is greater than B, so A > B. Verification / Alternative check:We can do a rough check without full decimal computation. Since 32^1 = 32 and 32^2 = 1024, while 32^3 = 32768, the number 1875 lies between 32^2 and 32^3. That means A is between 2 and 3. By contrast, for base 243, we have 243^1 = 243 and 243^2 = 59049, with 2187 lying between 243^1 and 243^2, but much closer to 243 than to 59049. Therefore B is clearly a number between 1 and 2 and closer to 1, which supports the conclusion that A > B. Why Other Options Are Wrong:The option A B is correct. Common Pitfalls:Some learners try to convert everything to base 10 and then fall into rounding errors. Others mis-handle the property log_(a^k) (a^m) = m / k or factorize the numbers incorrectly. Another common mistake is to rely on intuition instead of performing a systematic comparison. Using prime factorization and simple inequalities avoids these issues. Final Answer:The correct relationship is A > B, so A is greater than B.

Question 3

The value of $\log_{343} 7$ is

Accepted Answer

Concept & Formula When the base of a logarithm is larger than the argument, expect a fractional exponent. Use the base change or exponent conversion rule: \log_a b = x \implies a^x = b Step-by-Step Solution • Given: Evaluate \log_{343} 7. • Calculation: Let the unknown value be x. \log_{343} 7 = x • Convert the logarithmic equation into an exponential equation: 343^x = 7 • Recognize that 343 is a perfect cube of 7, meaning 343 = 7^3. Substitute this into the equation: (7^3)^x = 7^1 7^{3x} = 7^1 • Since the bases are identical, equate the exponents: 3x = 1 x = (1/3) Exam Strategy & Shortcut If you memorize cubes up to 10, you will instantly spot that 7^3 = 343. Since the logarithm asks "what power of 343 gives 7", taking the cube root (power of 1/3) is the immediate mental answer. Common Pitfall Students often answer with 3 instead of 1/3, confusing \log_{343} 7 with \log_7 343. Always pay attention to which number is the base and which is the argument. Final Answer Therefore, the correct answer is (1/3).

Question 4

The value of $\log_5 \frac{(125)(625)}{25}$ is equal to

Accepted Answer

Concept & Formula Simplify the argument of the logarithm first by expressing all terms with a common base, then apply the logarithm power rule: \log_a (a^n) = n Step-by-Step Solution • Given: Evaluate \log_5 ((125)(625)/25). • Calculation: First, simplify the fraction. Express all numbers as powers of 5: 125 = 5^3 625 = 5^4 25 = 5^2 • Substitute these into the fraction: (5^3 \cdot 5^4/5^2) • Apply the laws of exponents (x^a \cdot x^b = x^{a+b} and (x^a/x^b) = x^{a-b}): \frac{5^{3+4}}{5^2} = (5^7/5^2) = 5^{7-2} = 5^5 • Now, substitute this simplified value back into the original logarithm: \log_5 (5^5) • The base and the logarithm cancel out, leaving the exponent: 5 Exam Strategy & Shortcut Using logarithmic addition/subtraction rules directly is slower. Simplifying the exponential terms first as powers of 5 (3 + 4 - 2 = 5) gives the final answer in a single mental step. Common Pitfall Multiplying 125 × 625 manually creates unnecessarily large numbers (78125), wasting precious exam time and increasing the chance of arithmetic errors. Always use exponent rules. Final Answer Therefore, the correct answer is 5.

Question 5

The value of $\log_{\sqrt{2}} 32$ is

Accepted Answer

Concept & Formula When the base is a surd (root), convert the logarithmic equation to an exponential form to easily compare the powers of a common integer base. \log_a b = x \implies a^x = b Step-by-Step Solution • Given: Evaluate \log_{\sqrt{2}} 32. Let this expression equal x. \log_{\sqrt{2}} 32 = x • Calculation: Rewrite the expression in exponential form: (\sqrt{2})^x = 32 • Express both sides using the common base of 2. We know \sqrt{2} = 2^{1/2} and 32 = 2^5: (2^{1/2})^x = 2^5 • Apply the power of a power rule (a^m)^n = a^{m \cdot n}: 2^{x/2} = 2^5 • Since the bases are identical, equate the exponents: (x/2) = 5 x = 10 Exam Strategy & Shortcut Think of it this way: It takes 5 twos to make 32 (2^5 = 32). Since the base is \sqrt{2}, you need exactly twice as many to reach the same value. 5 × 2 = 10. Common Pitfall Students often write the answer as 5/2, incorrectly dividing the power of 32 (5) by 2 instead of multiplying. Remember that a smaller base requires a larger exponent. Final Answer Therefore, the correct answer is 10.

Question 6

Determine the value of $\log_{3\sqrt{2}} \left( \frac{1}{18} \right)$.

Accepted Answer

Concept & Formula To evaluate a logarithm with complex bases and fractional arguments, set the expression to a variable, convert to exponential form, and isolate the power. \log_a b = x \implies a^x = b Step-by-Step Solution • Given: Evaluate \log_{3\sqrt{2}} ( (1/18) ). • Calculation: Let the value be x. Convert to an exponential equation: (3\sqrt{2})^x = (1/18) • Square the base (3\sqrt{2}) to see its relationship with 18: (3\sqrt{2})^2 = 3^2 \cdot (\sqrt{2})^2 = 9 \cdot 2 = 18 • Rewrite the right side of the equation ((1/18)) using negative exponents: (1/18) = 18^{-1} • Substitute 18 with (3\sqrt{2})^2: 18^{-1} = ((3\sqrt{2})^2)^{-1} = (3\sqrt{2})^{-2} • Now, bring it back to our original equation: (3\sqrt{2})^x = (3\sqrt{2})^{-2} • Equate the exponents: x = -2 Exam Strategy & Shortcut Observe the relationship between 3\sqrt{2} and 18. Squaring 3 gives 9, and squaring \sqrt{2} gives 2. 9 × 2 = 18. Because the argument is 1/18, the exponent must be negative. Answer is -2. Common Pitfall Getting intimidated by the 3\sqrt{2} base. Don't panic—square it first to see if it relates to the argument. Most aptitude questions are designed with perfect squares or cubes in mind. Final Answer Therefore, the correct answer is -2.

Question 7

The value of $\log_{10} (.0001)$ is

Accepted Answer

Concept & Formula To evaluate logarithms involving decimals, convert the decimal into a fraction and then into a negative power of the base. \log_a (a^n) = n Step-by-Step Solution • Given: Evaluate \log_{10} (.0001). • Calculation: First, express the decimal .0001 as a fraction. Since there are four decimal places, it becomes: (1/10000) • Convert the denominator to a power of 10: (1/10^4) • Bring the power to the numerator by making the exponent negative: 10^{-4} • Substitute this back into the logarithmic expression: \log_{10} (10^{-4}) • Using the power rule, the base and log cancel out, leaving the exponent: -4 Exam Strategy & Shortcut For any base 10 logarithm of a decimal like 0.00...1, simply count the number of decimal places. 0.0001 has 4 decimal places. Therefore, it is 10^{-4}, and the logarithm is instantly -4. Common Pitfall A common error is missing the negative sign and answering 4, confusing .0001 with 10000. Remember that fractional values (between 0 and 1) yield negative logarithms when the base is greater than 1. Final Answer Therefore, the correct answer is -4.

Question 8

The value of $\log_{.01}(1000)$ is

Accepted Answer

Concept & Formula When both the base and the argument can be expressed as powers of the same number (e.g., 10), use the base and argument exponent rule: \log_{a^m} (a^n) = (n/m) Step-by-Step Solution • Given: Evaluate \log_{.01}(1000). • Calculation: Express both the base (0.01) and the argument (1000) as powers of 10. • For the base: 0.01 = (1/100) = 10^{-2}. • For the argument: 1000 = 10^3. • Substitute these into the logarithm: \log_{10^{-2}} (10^3) • Apply the property \log_{a^m} (b^n) = (n/m) \log_a b: (3/-2) \log_{10} 10 • Since \log_{10} 10 = 1, the expression simplifies to: -(3/2) Exam Strategy & Shortcut Don't write out the formulas. Just look at the powers of 10. The base .01 is -2 power. The argument 1000 is +3 power. The answer is always (power of argument) / (power of base) \rightarrow 3 / -2 = -3/2. Common Pitfall Students often invert the fraction, arriving at -2/3 instead of -3/2. Always remember it is the argument's exponent divided by the base's exponent, not the other way around. Final Answer Therefore, the correct answer is -(3/2).

Question 9

What is the value of $[\log_{10} (5 \log_{10} 100)]^2$?

Accepted Answer

Concept & Formula When dealing with nested mathematical operations, always solve from the innermost parentheses outward, strictly following the order of operations (BODMAS/PEMDAS). \log_a (a^n) = n Step-by-Step Solution • Given: Evaluate [\log_{10} (5 \log_{10} 100)]^2. • Calculation: Start with the innermost term: \log_{10} 100. • Since 100 = 10^2, we know that \log_{10} 10^2 = 2. • Substitute this back into the main expression: [\log_{10} (5 × 2)]^2 • Simplify the term inside the remaining parentheses: [\log_{10} (10)]^2 • Since \log_{10} 10 = 1, substitute this value: [1]^2 • Square the result: 1^2 = 1 Exam Strategy & Shortcut This question looks complex but evaluates in seconds mentally. Inner \log_{10} 100 is clearly 2. 5 × 2 is 10. \log_{10} 10 is 1. The square of 1 is 1. No pen or paper is needed. Common Pitfall A major mistake is applying the outer square prematurely to the internal elements, or confusing the square bracket with a coefficient (multiplying by 2 instead of squaring). Always resolve the logs completely before applying outer powers. Final Answer Therefore, the correct answer is 1.

Question 10

If $\log_3 x = -2$, then $x$ is equal to

Accepted Answer

Concept & Formula To find an unknown argument in a logarithmic equation, convert the logarithm into its exponential form. \log_a b = x \implies a^x = b A negative exponent indicates the reciprocal of the base raised to the positive exponent. a^{-n} = (1/a^n) Step-by-Step Solution • Given: The equation is \log_3 x = -2. • Calculation: Convert this logarithmic equation to an exponential equation using the base 3: 3^{-2} = x • Apply the negative exponent rule to rewrite 3^{-2}: x = (1/3^2) • Evaluate 3^2: x = (1/9) Exam Strategy & Shortcut Whenever you see a negative result for a logarithm, it immediately means the argument is a fraction (less than 1) if the base is greater than 1. You know 3^2 = 9, so the negative sign just flips it to 1/9. You can eliminate options (a), (b), and (c) instantly because the argument of a logarithm (x) must always be positive. Common Pitfall A very common mistake is multiplying the base by the exponent to get -6, or raising it normally and then adding a negative sign to get -9. Always remember that a negative exponent creates a fraction, not a negative number. Final Answer Therefore, the correct answer is (1/9).

Question 11

If $\log_8 p = 25$ and $\log_2 q = 5$, then

Accepted Answer

Concept & Formula To relate variables from two different logarithmic equations, convert both to exponential form and express them using a common prime base so their exponents can be directly compared. \log_a b = x \implies a^x = b Step-by-Step Solution • Given: Two equations: \log_8 p = 25 and \log_2 q = 5. • Calculation: Convert both equations to exponential form: Equation 1: 8^{25} = p Equation 2: 2^5 = q • Notice that the bases 8 and 2 are related. Convert the base 8 in Equation 1 to base 2: (2^3)^{25} = p • Simplify the exponents using (a^m)^n = a^{m \cdot n}: 2^{75} = p • Now look at Equation 2 (2^5 = q). We need to raise this equation to a power that turns 2^5 into 2^{75} to match p. • Since 75 \div 5 = 15, raise both sides of Equation 2 to the power of 15: (2^5)^{15} = q^{15} 2^{75} = q^{15} • Since both p and q^{15} are equal to 2^{75}, we can equate them: p = q^{15} Exam Strategy & Shortcut Observe the bases: 8 is 2^3. This means p grows 3 times faster in the exponent. 3 × 25 = 75. q is at exponent 5. How many times does 5 go into 75? 15 times. Therefore, q must be raised to the 15th power to catch up to p. Common Pitfall A typical mistake is trying to divide the logarithmic equations directly without converting to exponential form or aligning the bases, leading to incorrect algebraic relationships like p = q^5. Final Answer Therefore, the correct answer is p = q^{15}.

Question 12

If $\log_{10000} x = -\frac{1}{4}$, then $x$ is equal to

Accepted Answer

Concept & Formula The fundamental definition of a logarithm connects it directly to exponents. If a logarithm is written as \log_a b = c, it can be converted to its exponential form: b = a^c Step-by-Step Solution • Given: \log_{10000} x = -(1/4) • Conversion: Convert the logarithmic equation into its equivalent exponential form. x = (10000)^{-(1/4)} • Prime Factorization: Express the base 10000 as a power of 10. We know that 10000 = 10^4. x = (10^4)^{-(1/4)} • Exponent Rules: Apply the power of a power property, which states (a^m)^n = a^{mn}. x = 10^{4 × -(1/4)} • Simplification: The 4 in the numerator and denominator cancel out. x = 10^{-1} • Final Value: Use the negative exponent rule a^{-n} = (1/a^n). x = (1/10) Exam Strategy & Shortcut When dealing with large powers of 10 and fractional exponents, always rewrite the large base as a power of 10 first. Recognizing 10000 as 10^4 mentally allows you to immediately cancel the 4 in the denominator of the exponent -(1/4), jumping straight to 10^{-1} and skipping unnecessary written steps. Common Pitfall A frequent mistake is interpreting a negative exponent as making the entire number negative (e.g., thinking the answer is -10 or -(1/10000)). Remember that negative exponents dictate reciprocals, not negative values. Final Answer Therefore, the correct answer is (1/10).

Question 13

If $\log_x y = 100$ and $\log_2 x = 10$, then the value of $y$ is

Accepted Answer

Concept & Formula This problem involves solving a system of sequential logarithmic equations. You must use the fundamental definition of logarithms twice: \log_a b = c \implies b = a^c Additionally, you will apply the exponent rule for raising a power to a power: (a^m)^n = a^{mn}. Step-by-Step Solution • Given: Two equations: \log_x y = 100 and \log_2 x = 10. • Solving the First Variable: Start with the equation that has a known base to find x. \log_2 x = 10 • Conversion: Convert this to exponential form. x = 2^{10} • Substituting into Second Equation: Now look at the other given equation: \log_x y = 100. Convert it to exponential form to isolate y. y = x^{100} • Combining Values: Substitute the value of x we found (2^{10}) into this new equation. y = (2^{10})^{100} • Simplification: Multiply the exponents together using the power rule. y = 2^{10 × 100} y = 2^{1000} Exam Strategy & Shortcut You can solve this rapidly using the change of base property or chaining. Recognize that y = x^{100} and x = 2^{10}. Mental substitution (2^{10})^{100} immediately yields 2^{1000}. No intermediate calculations of 2^{10} (1024) are necessary since the options remain in exponential form. Common Pitfall A common mistake is adding the exponents instead of multiplying them when evaluating (2^{10})^{100}, leading to the incorrect answer 2^{110}. Another trap is simply multiplying 100 and 10 as coefficients, completely misapplying logarithm rules. Always strictly follow the power of a power rule. Final Answer Therefore, the correct answer is 2^{1000}.

Question 14

The value of $\log_{(-\frac{1}{3})} 81$ is equal to

Accepted Answer

Concept & Formula To evaluate a logarithm, we rely on its relationship with exponents. The foundational equation converting a logarithm to exponential form is: x = \log_a b \implies a^x = b By setting the given logarithmic expression equal to a variable, we can solve the resulting exponential equation by matching bases. Step-by-Step Solution • Given: \log_{(-(1/3))} 81 • Set up the equation: Let the value of the expression be x. \log_{(-(1/3))} 81 = x • Convert to exponential form: (-(1/3))^x = 81 • Express both sides with the same base: We know that 81 is a power of 3 (3^4 = 81). We can also write -(1/3) as a negative power. • Notice that (-(1/3))^{-4} = (-3)^4 = 81. • Match the exponents: Since (-(1/3))^{-4} = 81, we can directly deduce the value of x. x = -4 Exam Strategy & Shortcut Whenever you have a fraction as a base and an integer as the argument, expect the answer to be negative. Recognizing that 3^4 = 81 immediately tells you that the exponent must involve a 4. Since the base is flipped (a fraction (1/3)), the exponent must be negative to flip it back to a whole number 3. Thus, -4 is the logical quick answer. Common Pitfall A common mistake is ignoring the negative sign on the fraction or simply guessing 4 because 3^4 = 81. However, (-(1/3))^4 = (1/81), not 81. Always verify the sign of your resulting exponent by plugging it back into the base. Final Answer Therefore, the correct answer is -4.

Question 15

$\log_{10} (10 	imes 10^2 	imes 10^3 	imes ...... 	imes 10^9)$ is

Accepted Answer

Concept & Formula This problem combines the product rule of exponents with the fundamental property of base-10 logarithms. When multiplying expressions with the same base, you add their exponents: a^m × a^n = a^{m+n} Additionally, the logarithm property states: \log_{10} (10^k) = k Step-by-Step Solution • Given: \log_{10} (10 × 10^2 × 10^3 × \dots × 10^9) • Simplify the Argument: Use the exponent rule to combine the terms inside the parenthesis. The base is 10 for all terms, so we sum the exponents from 1 to 9. 10^{1 + 2 + 3 + \dots + 9} • Calculate the Sum: Use the formula for the sum of the first n natural numbers, which is (n(n+1)/2), where n = 9. Sum = (9(9+1)/2) = (9 × 10/2) = 45 • Substitute the Sum: The argument simplifies to 10^{45}. • Evaluate the Logarithm: Substitute this back into the original logarithmic expression. \log_{10} (10^{45}) • Final Calculation: Applying the rule \log_a (a^x) = x, the log and the base 10 cancel each other out. 45 Exam Strategy & Shortcut Recognize the pattern immediately: this is just the sum of numbers from 1 to 9. Knowing that the sum of 1 through 9 is 45 (a common arithmetic fact) allows you to bypass writing out any steps. The question instantly reduces to \log_{10}(10^{45}) = 45 in your head, taking less than 5 seconds. Common Pitfall A common error is confusing the product rule of exponents with multiplication, leading students to multiply the exponents (e.g., 1 × 2 × 3 \dots × 9 = 9!) instead of adding them. Remember that multiplying same bases means adding their powers. Final Answer Therefore, the correct answer is 45.

Question 16

$\log_{10} \log_{10} \log_{10} (10^{10^{10}})$ is equal to

Accepted Answer

Concept & Formula This question deals with nested logarithms operating on a power tower (an iterated exponentiation). The key property to apply repeatedly is: \log_a (a^x) = x This rule allows the outermost exponent to "drop down" when the base of the logarithm matches the base of the exponential term. Step-by-Step Solution • Given: \log_{10} [ \log_{10} [ \log_{10} (10^{10^{10}}) ] ] • Evaluate Innermost Logarithm: Apply the rule \log_{10}(10^x) = x to the innermost logarithm. Here, the exponent x is the upper part of the tower, 10^{10}. \log_{10} (10^{10^{10}}) = 10^{10} • Substitute and Evaluate Second Logarithm: Now, the expression becomes: \log_{10} [ \log_{10} (10^{10}) ] • Apply the same rule again. Ask: "What is the exponent on the base 10?" The exponent is 10. \log_{10} (10^{10}) = 10 • Substitute and Evaluate Outermost Logarithm: The expression is now reduced to a single logarithm: \log_{10} (10) • Final Calculation: Since 10^1 = 10, this evaluates to 1. 1 Exam Strategy & Shortcut Each time you take \log_{10} of a power of 10, you simply chop off the bottom "10" of the exponent tower. There are three \log_{10} operations and three "10"s in the exponent tower 10^{10^{10}}. Operation 1 knocks out the base 10 leaving 10^{10}. Operation 2 knocks out the next 10 leaving 10. Operation 3 knocks out the final 10 leaving an implied exponent of 1. Therefore, they cancel out perfectly down to 1. Common Pitfall Students frequently misinterpret 10^{10^{10}} as (10^{10})^{10}, which would be 10^{100}. A power tower is always evaluated from the top down, so 10^{10^{10}} means 10^{(10^{10})}. Misinterpreting this can cause you to apply the log rules incorrectly and get stuck with an extra factor. Final Answer Therefore, the correct answer is 1.

Question 17

The value of $\log_2 \log_2 \log_3 \log_3 27^3$ is

Accepted Answer

Concept & Formula This problem tests the sequential evaluation of nested logarithms, working from the rightmost (innermost) term to the left. The essential properties used are: \log_a (b^n) = n \log_a b \log_a a = 1 a^0 = 1 \implies \log_a 1 = 0 Step-by-Step Solution • Given: \log_2 \log_2 \log_3 \log_3 (27^3) • Simplify the Innermost Argument: First, express 27 as a power of 3. 27^3 = (3^3)^3 • Apply the power of a power rule (a^m)^n = a^{mn}. (3^3)^3 = 3^9 • Evaluate the First Logarithm (Rightmost): \log_3 (3^9) = 9 • Substitute and Evaluate the Second Logarithm: Replace the evaluated part in the main expression. \log_2 \log_2 \log_3 (9) • Since 9 = 3^2, evaluate \log_3 9. \log_3 (3^2) = 2 • Substitute and Evaluate the Third Logarithm: \log_2 \log_2 (2) • Since 2^1 = 2, evaluate \log_2 2. \log_2 2 = 1 • Evaluate the Final Logarithm (Leftmost): \log_2 (1) • Since any non-zero base raised to the power of 0 is 1, \log_a 1 = 0. 0 Exam Strategy & Shortcut Solve strictly right-to-left. Recognize that 27^3 is 3^9. Mentally step through the chain: 1. \log_3 of 3^9 is 9. 2. \log_3 of 9 is 2. 3. \log_2 of 2 is 1. 4. \log_2 of 1 is 0. This structured mental chain bypasses the need for any complex written algebra. Common Pitfall A common trap is incorrectly calculating the initial 27^3 term by doing something like 3 × 3 = 9 or confusing it with 27 × 3 = 81. Ensure you use the correct exponent rule: (3^3)^3 = 3^9. Another mistake is stopping at the value of 1, forgetting the outermost \log_2 which turns the 1 into a 0. Final Answer Therefore, the correct answer is 0.

Question 18

If $a^x = b^y$, then

Accepted Answer

Concept & Formula: This problem requires converting an exponential equation into a logarithmic relationship. The key insight is taking the logarithm of both sides of an equation to bring variables out of the exponents. The primary rule used is the power rule of logarithms: \log(m^n) = n \cdot \log(m) Step-by-Step Solution: Given: a^x = b^y To isolate the exponents x and y, we take the logarithm (to any common base, typically base 10) on both sides of the equation. Take log on both sides: \log(a^x) = \log(b^y) Apply the power rule \log(m^n) = n \cdot \log(m) to bring the exponents to the front: x \cdot \log(a) = y \cdot \log(b) Now, we need to match the format of the options, which represent ratios. Let's isolate the ratio of the logarithms. Divide both sides by x \cdot \log(b): (\log(a)/\log(b)) = (y/x) Exam Strategy & Shortcut: When equating exponents with different bases like a^x = b^y, immediately know that their logarithmic ratio is inversely proportional to their exponent ratio. Thus, \log a over \log b will always equal y over x. Recognizing this pattern allows you to bypass the intermediate steps and directly select the correct ratio in seconds. Common Pitfall: A very common mistake is confusing the inverse relationship when cross-multiplying, leading students to choose (\log a/\log b) = (x/y). Always double-check your algebraic rearrangement: if x is multiplied by \log a, then x must end up in the denominator of the opposite side when setting up the fraction. Final Answer: Therefore, the correct answer is (\log a/\log b) = (y/x).

Question 19

$\log_{10} \frac{26}{51} + \log_{10} \frac{119}{91} - \log_{10} \frac{13}{32} - \log_{10} \frac{64}{39}$ is equal to

Accepted Answer

Concept & Strategy: This problem tests the application of the product and quotient rules of logarithms to combine a string of logarithmic terms into a single expression. Addition of logs translates to multiplication of their arguments, while subtraction translates to division (or multiplication by the reciprocal). \log(A) + \log(B) - \log(C) = \log((A \cdot B/C)) Step-by-Step Solution: Given: \log_{10} ((26/51)) + \log_{10} ((119/91)) - \log_{10} ((13/32)) - \log_{10} ((64/39)) Using the log properties, we can combine all terms into a single base-10 logarithm. Terms being added are multiplied in the numerator; terms being subtracted are divided (which means we multiply by their reciprocals). Combine the arguments: \log_{10} ( (26/51) × (119/91) × (32/13) × (39/64) ) Now, break down all numbers into prime factors to easily cancel terms out. Numerator terms: 26 = 2 × 13 119 = 7 × 17 32 = 2^5 39 = 3 × 13 Combined Numerator = 2 × 13 × 7 × 17 × 2^5 × 3 × 13 = 2^6 × 3 × 7 × 13^2 × 17 Denominator terms: 51 = 3 × 17 91 = 7 × 13 13 = 13 64 = 2^6 Combined Denominator = 3 × 17 × 7 × 13 × 13 × 2^6 = 2^6 × 3 × 7 × 13^2 × 17 Notice that the combined numerator is exactly identical to the combined denominator. Therefore, the entire fraction simplifies to 1. Substitute back into the log: \log_{10} (1) Since the logarithm of 1 to any base is always 0: \log_{10} (1) = 0 Exam Strategy & Shortcut: When faced with a long string of fractional log additions and subtractions, write one giant fraction immediately. Flip the fractions that have a minus sign in front of their log and multiply everything. Cancel common factors aggressively before doing any actual multiplication. In competitive exams, these fractions are specifically designed to perfectly cancel out to 1, 10, or 100. Common Pitfall: The most frequent error is mishandling the subtracted logs. Students sometimes subtract the fractions directly or forget to invert the fractions before multiplying. Remember: -\log(a/b) is equivalent to +\log(b/a). Always flip the subtracted terms before combining. Final Answer: Therefore, the correct answer is 0.

Question 20

The value of $\left(\frac{1}{3}\log_{10} 125 - 2\log_{10} 4 + \log_{10} 32\right)$ is

Accepted Answer

Concept & Formula: This problem requires consolidating multiple logarithmic terms into a single logarithm using the power, product, and quotient rules. The Power Rule: \log_a(x^n) = n \cdot \log_a(x) The Product Rule: \log_a(xy) = \log_a(x) + \log_a(y) The Quotient Rule: \log_a((x/y)) = \log_a(x) - \log_a(y) Step-by-Step Solution: Let's simplify each term individually using the power rule to bring the coefficient into the exponent. First term: (1/3)\log_{10}(125) Since 125 = 5^3, we have: (1/3)\log_{10}(5^3) = \log_{10}((5^3)^{1/3}) = \log_{10}(5^1) = \log_{10}(5) Second term: 2\log_{10}(4) Bring the 2 up as a power: \log_{10}(4^2) = \log_{10}(16) Third term remains as is: \log_{10}(32) Now substitute these simplified terms back into the original expression: \log_{10}(5) - \log_{10}(16) + \log_{10}(32) Apply the product and quotient rules to combine them into a single log: \log_{10}((5 × 32/16)) Simplify the fraction: (5 × 32/16) = 5 × 2 = 10 So the expression becomes: \log_{10}(10) Since \log_a(a) = 1, we get: \log_{10}(10) = 1 Exam Strategy & Shortcut: Instead of combining into a single large number, convert everything to prime bases immediately. 125 = 5^3, 4 = 2^2, 32 = 2^5. Expression = (1/3) \cdot 3\log_{10}(5) - 2 \cdot 2\log_{10}(2) + 5\log_{10}(2) = \log_{10}(5) - 4\log_{10}(2) + 5\log_{10}(2) = \log_{10}(5) + \log_{10}(2) = \log_{10}(10) = 1. This avoids fraction multiplication entirely. Common Pitfall: A frequent mistake is mishandling the power rule with fractional coefficients. Students might incorrectly evaluate (1/3)\log(125) by dividing the entire log value rather than applying the cube root to the argument 125. Always apply the exponent to the argument first. Final Answer: Therefore, the correct answer is 1.

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