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 $\log_{\sqrt{2}} 32$ 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 5

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 6

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 7

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 8

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 9

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 10

$\log_{10} \log_{10} \log_{10} (10^{10^{10}})$ 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 11

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 12

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.

Question 13

The value of $\log_{10} 1\frac{1}{2} + \log_{10} 1\frac{1}{3} + \cdots$ up to $198$ terms is equal to

Accepted Answer

Concept & Strategy: This question relies on recognizing a telescoping series within a logarithmic addition. When adding logarithms of fractions, the product rule allows you to multiply all the fractions together, leading to massive cancellation. The Product Rule: \log(a) + \log(b) + \log(c) = \log(a \cdot b \cdot c) Step-by-Step Solution: First, convert all the mixed fractions into improper fractions to see the pattern clearly. Term 1: 1(1/2) = (3/2) Term 2: 1(1/3) = (4/3) Term 3: 1(1/4) = (5/4) ...and so on. Let's determine the 198th term. Notice the pattern in the denominator of the n-th term: it is n + 1. For the 198th term, the denominator is 198 + 1 = 199. The numerator is always 1 greater than the denominator, so the 198th numerator is 200. Thus, the 198th term is (200/199). The expression is: \log_{10}((3/2)) + \log_{10}((4/3)) + \cdots + \log_{10}((200/199)) Use the product rule to combine all terms into a single logarithm: \log_{10}( (3/2) × (4/3) × (5/4) × \cdots × (200/199) ) Observe the telescoping cancellation. The numerator of each fraction cancels perfectly with the denominator of the next fraction. The 3s cancel, the 4s cancel, all the way up to 199. You are left with only the final numerator and the first denominator: \log_{10}((200/2)) Simplify the argument: \log_{10}(100) Evaluate the logarithm: \log_{10}(10^2) = 2 Exam Strategy & Shortcut: Whenever you see a continuous series of "log(1 + 1/n)" additions, instantly know it's a telescoping product. Write down only the first term's denominator and the last term's numerator. Skip writing out the middle terms completely. 200 / 2 = 100. The log of 100 is 2. You can solve this mentally in 5 seconds. Common Pitfall: The main trap is incorrectly identifying the final term of the series. Students often see "up to 198 terms" and mistakenly use 198 in the final fraction (e.g., 199/198), failing to account for the sequence starting with a denominator of 2. Always map n = 1 to the first term to find the correct algebraic shift. Final Answer: Therefore, the correct answer is 2.

Question 14

What is the value of the following expression?  $\log \left(\frac{9}{14}\right) - \log \left(\frac{15}{16}\right) + \log \left(\frac{35}{24}\right)$

Accepted Answer

Concept & Strategy: This problem tests your ability to combine multiple logarithms using product and quotient rules. Addition of logs means multiplying the arguments, while subtraction means dividing (which is equivalent to multiplying by the reciprocal). \log(A) - \log(B) + \log(C) = \log(A \cdot (1/B) \cdot C) Step-by-Step Solution: Given: \log ((9/14)) - \log ((15/16)) + \log ((35/24)) Combine the logarithmic terms into a single log expression. The subtracted term's fraction must be inverted. \log ( (9/14) × (16/15) × (35/24) ) Now, simplify the large fraction by factoring it into primes to cancel common terms out between the numerator and denominator. Numerator: 9 × 16 × 35 = (3^2) × (2^4) × (5 × 7) = 2^4 × 3^2 × 5 × 7 Denominator: 14 × 15 × 24 = (2 × 7) × (3 × 5) × (2^3 × 3) = 2^4 × 3^2 × 5 × 7 Compare the expanded numerator and denominator. They are perfectly identical. Therefore, the fraction simplifies to 1. Substitute back into the log expression: \log(1) The logarithm of 1 to any base is always 0. \log(1) = 0 Exam Strategy & Shortcut: Do not multiply large numbers like 9 × 16 × 35. Always break them into prime factors immediately. Cancel the 2s, 3s, 5s, and 7s out visually on your paper. In most aptitude exams, expressions constructed like this are heavily rigged to cancel out to exactly 1 or 10. Common Pitfall: Students often subtract fractions directly rather than applying logarithmic rules, or they forget to flip the fraction that is being subtracted. Always remember that -\log(x/y) becomes +\log(y/x) before you multiply the arguments. Final Answer: Therefore, the correct answer is 0.

Question 15

If $\log_{10} (x^2 - 6x + 10) = 0$, then the value of $x$ is

Accepted Answer

Concept & Logic: This problem combines basic logarithmic definitions with solving a quadratic equation. The fundamental definition of a logarithm converts it into an exponential equation. Logarithm to Exponential Conversion: If \log_b(A) = C , then b^C = A Step-by-Step Solution: Given: \log_{10}(x^2 - 6x + 10) = 0 Convert this logarithmic equation into its equivalent exponential form. The base is 10, the exponent is 0, and the result is the quadratic expression: 10^0 = x^2 - 6x + 10 Since any non-zero number raised to the power of 0 is 1: 1 = x^2 - 6x + 10 Rearrange the equation to set it to zero (standard quadratic form): x^2 - 6x + 10 - 1 = 0 x^2 - 6x + 9 = 0 Factor the quadratic equation. We need two numbers that multiply to 9 and add to -6. Those numbers are -3 and -3. (x - 3)(x - 3) = 0 (x - 3)^2 = 0 Take the square root of both sides: x - 3 = 0 x = 3 Exam Strategy & Shortcut: Whenever you see a log equation equal to zero, you instantly know the argument (what's inside the parentheses) must equal 1. So, x^2 - 6x + 10 = 1 \Rightarrow x^2 - 6x + 9 = 0. Recognizing x^2 - 6x + 9 as a perfect square (x-3)^2 allows you to solve it mentally in seconds without writing down the intermediate steps. Common Pitfall: A frequent error is forgetting that 10^0 = 1 and mistakenly assuming 10^0 = 0, leading to the equation x^2 - 6x + 10 = 0, which has complex roots and will derail the entire solving process. Final Answer: Therefore, the correct answer is 3.

Question 16

If $\log_{10} x + \log_{10} y = 3$ and $\log_{10} x - \log_{10} y = 1$, then $x$ and $y$ are respectively

Accepted Answer

Concept & Strategy: This is a system of linear equations disguised as a logarithms problem. Treat \log_{10}(x) and \log_{10}(y) as individual variables (like A and B) and solve using the standard elimination method. Step-by-Step Solution: Let A = \log_{10}(x) and B = \log_{10}(y). The given equations become: Equation 1: A + B = 3 Equation 2: A - B = 1 Use the elimination method. Add Equation 1 and Equation 2 together: (A + B) + (A - B) = 3 + 1 2A = 4 A = 2 Now substitute A = 2 back into Equation 1 to find B: 2 + B = 3 B = 1 Now, substitute the logarithmic definitions back for A and B: \log_{10}(x) = 2 \log_{10}(y) = 1 Convert to exponential form to solve for x and y: x = 10^2 = 100 y = 10^1 = 10 So, x and y are respectively 100 and 10. Exam Strategy & Shortcut: You don't even need to solve the linear equations. Use the product and quotient rules directly on the values. A + B = 3 means \log_{10}(xy) = 3 \Rightarrow xy = 1000. A - B = 1 means \log_{10}(x/y) = 1 \Rightarrow x/y = 10. Look at the options. (a) 10 × 100 = 1000 (Pass), 10/100 = 0.1 (Fail) (b) 100 × 10 = 1000 (Pass), 100/10 = 10 (Pass) Option (b) is the correct answer. Common Pitfall: The main pitfall is swapping the order of x and y in the final answer. The question asks for "x and y respectively". Since x = 100 and y = 10, selecting Option (a) "10 and 100" is a careless mistake that costs full marks despite doing the math correctly. Final Answer: Therefore, the correct answer is 100 and 10.

Question 17

The value of $(\log_9 27 + \log_8 32)$ is

Accepted Answer

Concept & Formula: This problem requires evaluating two separate logarithms by expressing both their bases and arguments as powers of a common prime number. The fundamental rule to use here is the base-power and argument-power property: \log_{a^k} (b^n) = (n/k) \cdot \log_a(b) Step-by-Step Solution: Let's evaluate each logarithmic term individually before adding them. First term: \log_9(27) Express both 9 and 27 as powers of the prime base 3: 9 = 3^2 27 = 3^3 Rewrite the logarithm: \log_{3^2}(3^3) Apply the power rule to pull out the exponents as a fraction: (3/2) \cdot \log_3(3) Since \log_3(3) = 1, the term simplifies to: (3/2) Second term: \log_8(32) Express both 8 and 32 as powers of the prime base 2: 8 = 2^3 32 = 2^5 Rewrite the logarithm: \log_{2^3}(2^5) Apply the power rule: (5/3) \cdot \log_2(2) Since \log_2(2) = 1, the term simplifies to: (5/3) Combine the terms: Now add the simplified values of the two terms together: (3/2) + (5/3) To add these fractions, find a common denominator, which is 6: ((3 \cdot 3/2 \cdot 3)) + ((5 \cdot 2/3 \cdot 2)) (9/6) + (10/6) (19/6) Exam Strategy & Shortcut: For any log where the base and argument are powers of the same number, the answer is just the ratio of their exponents: (power of argument) / (power of base). 27 is 3^3, 9 is 3^2 \Rightarrow 3/2. 32 is 2^5, 8 is 2^3 \Rightarrow 5/3. You can skip writing down the log notations entirely and jump straight to adding the fractions 3/2 + 5/3. Common Pitfall: A very common mistake is flipping the fraction when applying the exponent rule, writing \log_9(27) as 2/3 instead of 3/2. Remember the rule: the exponent of the argument goes in the numerator (top), and the exponent of the base goes in the denominator (bottom). Final Answer: Therefore, the correct answer is (19/6).

Question 18

$(\log_5 3) 	imes (\log_3 625)$ equals

Accepted Answer

Concept & Formula: This problem is solved efficiently using the "Chain Rule" of logarithms, which is derived from the Change of Base formula. The Change of Base formula allows us to rewrite any logarithm as a fraction: \log_a(b) = (\log(b)/\log(a)) By multiplying two such logarithms, we get the Chain Rule property: \log_a(b) \cdot \log_b(c) = \log_a(c) Step-by-Step Solution: Given expression: (\log_5 3) × (\log_3 625) Using the Change of Base formula, convert both logarithms to a common base (like base 10): First term: \log_5(3) = (\log(3)/\log(5)) Second term: \log_3(625) = (\log(625)/\log(3)) Multiply the two fractions together: (\log(3)/\log(5)) × (\log(625)/\log(3)) Notice that \log(3) appears in both the numerator and the denominator, so they cancel each other out completely: (\log(625)/\log(5)) Now, convert this fraction back into a single logarithm using the reverse of the Change of Base formula: \log_5(625) To evaluate this, express 625 as a power of the base 5: 625 = 5 × 5 × 5 × 5 = 5^4 Substitute this back: \log_5(5^4) Using the power rule, this simplifies to just the exponent: 4 Exam Strategy & Shortcut: You do not need to convert to fractions on paper. The Chain Rule allows you to visually "cancel" the diagonal matching bases and arguments. In (\log_5 3) × (\log_3 625), the argument 3 and base 3 cancel instantly, leaving you with \log_5(625). Since 5^4 = 625, the answer is simply 4. This takes 5 seconds mentally. Common Pitfall: A common mistake is attempting to multiply the arguments directly (3 × 625) while keeping the bases separate, which violates all rules of logarithms. Logarithms cannot be multiplied internally like standard algebraic terms unless you are applying the change of base property. Final Answer: Therefore, the correct answer is 4.

Question 19

$(\log_5 5) (\log_4 9) (\log_3 2)$ is equal to

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Concept & Formula: This problem involves simplifying individual logarithmic terms using the base-power property and then combining them using the reciprocal/chain rules of logarithms. Base-Power Property: \log_{a^n}(b^n) = \log_a(b) Reciprocal Property: \log_a(b) \cdot \log_b(a) = 1 Step-by-Step Solution: Given expression: (\log_5 5) \cdot (\log_4 9) \cdot (\log_3 2) Let's simplify each term one by one. First term: \log_5(5) The logarithm of any base to itself is always 1. \log_5(5) = 1 Second term: \log_4(9) Recognize that both the base and the argument are perfect squares. Express them as powers: 4 = 2^2 9 = 3^2 Substitute these into the term: \log_{2^2}(3^2) Apply the base-power property ((2/2) = 1), reducing the term to: \log_2(3) Third term remains as is: \log_3(2) Now, bring all the simplified terms back together into the original multiplication expression: 1 \cdot (\log_2 3) \cdot (\log_3 2) (\log_2 3) \cdot (\log_3 2) Use the reciprocal property of logarithms. Since the base of one is the argument of the other, their product is 1: \log_2(3) × (1/\log_2(3)) = 1 Exam Strategy & Shortcut: Whenever you see a logarithm where both the base and argument are raised to the same power, strip the powers immediately. \log_4(9) is instantly just \log_2(3). From there, you have a chain of \log_2(3) × \log_3(2). The bases and arguments criss-cross and cancel perfectly, leaving 1. The \log_5(5) is just a distractor that also equals 1. 1 × 1 = 1. Common Pitfall: A common trap is trying to convert everything into base 10 fractions right away without simplifying the perfect squares first. While (\log 9/\log 4) × (\log 2/\log 3) will eventually yield the right answer, it creates unnecessary and messy arithmetic steps that consume valuable exam time. Final Answer: Therefore, the correct answer is 1.

Question 20

If $\log_{12} 27 = a$, then $\log_6 16$ is

Accepted Answer

Concept & Strategy: This is an advanced algebraic manipulation problem. The standard strategy is to express everything in terms of a single, simple prime base (like base 2 or base 3). By breaking down both the "given" equation and the "target" expression into the same base, you can isolate the core logarithmic variable and substitute it. Change of Base Formula: \log_b(x) = (\log_k(x)/\log_k(b)) Step-by-Step Solution: Let's convert everything to base 2. Step 1: Simplify the Given Equation Given: a = \log_{12}(27) Change to base 2: a = (\log_2(27)/\log_2(12)) Break down 27 and 12 into prime factors: 27 = 3^3 12 = 2^2 × 3 Substitute these in and expand: a = (\log_2(3^3)/\log_2(2^2 \cdot 3)) a = (3 \cdot \log_2(3)/2 \cdot \log_2(2) + \log_2(3)) Since \log_2(2) = 1, we get: a = (3 \cdot \log_2(3)/2 + \log_2(3)) Step 2: Isolate \log_2(3) Let x = \log_2(3) for easier algebra. a = (3x/2 + x) Cross-multiply: a(2 + x) = 3x 2a + ax = 3x 2a = 3x - ax 2a = x(3 - a) x = (2a/3 - a) So, \log_2(3) = (2a/3 - a). Step 3: Simplify the Target Expression We need to find \log_6(16). Change to base 2: \log_6(16) = (\log_2(16)/\log_2(6)) We know 16 = 2^4 and 6 = 2 × 3: = (4 \cdot \log_2(2)/\log_2(2) + \log_2(3)) = (4/1 + \log_2(3)) Step 4: Substitute and Solve Substitute our value of \log_2(3) from Step 2 into the target expression: = (4/1 + (\frac{2a){3 - a})} Find a common denominator for the bottom: = (4/\frac{3 - a + 2a){3 - a}} = (4/\frac{3 + a){3 - a}} Multiply by the reciprocal: = 4 \cdot (3 - a/3 + a) = (4(3 - a)/3 + a) Exam Strategy & Shortcut: Problems involving variables and base changes take too long algebraically. Use dummy numbers if you get stuck. Since a is a specific value (\approx 1.326), and the options involve a, this indicates the algebraic relationship is absolute. However, mastering the algebraic isolation technique shown above is ultimately safer and often just as fast once you get accustomed to swapping out expressions for variables like x. Common Pitfall: A major pitfall is choosing the wrong base for the initial change of base. While base 10 or base 3 works, choosing base 2 is optimal here because 16 (in the target log) is a clean power of 2, simplifying the numerator to an integer 4 instantly. Choosing base 3 makes the final substitution slightly messier. Final Answer: Therefore, the correct answer is (4 (3 - a)/(3 + a)).

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