Question 1

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 2

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 3

If $\log_a m = x$, then $\log_{1/a} \left(\frac{1}{m}\right)$ equals

Accepted Answer

Concept & Formula: This problem deals with the manipulation of logarithmic bases and arguments when both are expressed as reciprocals (or negative powers). The relevant property for logarithms with exponents in both the base and argument is: \log_{a^k} (b^n) = (n/k) \cdot \log_a (b) Step-by-Step Solution: Given: \log_a (m) = x We need to find the value of \log_{1/a} ((1/m)). First, rewrite both the base and the argument as powers of their original variables. The reciprocal of a number is simply that number raised to the power of -1: (1/a) = a^{-1} (1/m) = m^{-1} Substitute these into the target expression: \log_{a^{-1}} (m^{-1}) Now, apply the exponent property of logarithms, bringing the exponent of the argument to the numerator, and the exponent of the base to the denominator: \log_{a^{-1}} (m^{-1}) = (-1/-1) \cdot \log_a (m) Simplify the fraction: 1 \cdot \log_a (m) = \log_a (m) Since we were given that \log_a (m) = x, substitute x back into the simplified expression: \log_a (m) = x Exam Strategy & Shortcut: A general rule of thumb for logarithms: if you invert BOTH the base and the argument, the value of the logarithm remains completely unchanged. Since both a and m are inverted to 1/a and 1/m, the negative signs cancel out instantly. You can immediately select x without writing down a single step. Common Pitfall: The most frequent mistake is pulling out a single negative sign from either the base or the argument, leading students to incorrectly choose -x. Remember that both the base and argument contribute a negative sign to the coefficient, which multiply to become positive. Final Answer: Therefore, the correct answer is x.

Question 4

If $\log_4 x + \log_2 x = 6$, then $x$ is equal to

Accepted Answer

Concept & Formula: This problem requires making the bases of the logarithms identical so they can be combined. The change of base formula or the property of base powers is essential here. Base Power Property: \log_{a^n}(x) = (1/n) \cdot \log_a(x) Step-by-Step Solution: Given: \log_4(x) + \log_2(x) = 6 First, convert \log_4(x) to base 2 so it matches the other term. Notice that 4 = 2^2. Using the base power property: \log_{2^2}(x) = (1/2) \cdot \log_2(x) Now substitute this back into the original equation: (1/2)\log_2(x) + \log_2(x) = 6 Factor out \log_2(x): \log_2(x) ((1/2) + 1) = 6 (3/2)\log_2(x) = 6 Isolate \log_2(x) by multiplying both sides by (2/3): \log_2(x) = 6 \cdot (2/3) \log_2(x) = 4 Convert the logarithmic equation to its exponential form to solve for x: x = 2^4 x = 16 Exam Strategy & Shortcut: Instead of algebraic manipulation, you can plug the options into the equation, starting from the easiest numbers. Let's try option (d) x = 16: \log_4(16) + \log_2(16) 4^2 = 16, so the first term is 2. 2^4 = 16, so the second term is 4. 2 + 4 = 6. This matches the equation perfectly. Option checking is much faster here. Common Pitfall: A common mistake is attempting to use the product rule directly (e.g., adding the arguments to get 2x or multiplying bases) when the bases are different. Logarithmic terms cannot be directly added or combined using the product rule unless their bases are exactly the same. Final Answer: Therefore, the correct answer is 16.

Question 5

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 6

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 7

If $\log_8 x + \log_8 \frac{1}{6} = \frac{1}{3}$, then the value of $x$ is

Accepted Answer

Concept & Formula: Similar to previous problems, this uses the product rule of logarithms to combine terms, and then evaluates the result using fractional exponents (roots). Product Rule: \log_a(M) + \log_a(N) = \log_a(M \cdot N) Fractional Exponent Rule: a^{1/n} = \sqrt[n]{a} Step-by-Step Solution: Given: \log_8(x) + \log_8((1/6)) = (1/3) Apply the product rule to combine the terms on the left, as they share the same base (8): \log_8(x \cdot (1/6)) = (1/3) \log_8((x/6)) = (1/3) Convert this logarithmic equation into its exponential equivalent: 8^{1/3} = (x/6) Evaluate the fractional exponent. Raising a number to the power of (1/3) is the same as taking its cube root: 8^{1/3} = \sqrt[3]{8} = 2 Substitute this back into the equation: 2 = (x/6) Multiply both sides by 6 to solve for x: x = 2 \cdot 6 x = 12 Exam Strategy & Shortcut: Memorize your basic cube roots (2^3 = 8, 3^3 = 27, etc.). The moment you see an equation equal to 1/3 with a base of 8, immediately recognize that the right side of the exponential equation will resolve to the clean integer 2. This simplifies the problem mentally to x/6 = 2, yielding 12 almost instantly. Common Pitfall: Students often struggle with fractional exponents and might mistakenly evaluate 8^{1/3} as 8/3, leading to messy fractions and an incorrect final answer. Always remember that a fractional exponent signifies a root, not division. Final Answer: Therefore, the correct answer is 12.

Question 8

If $\log_{10} 125 + \log_{10} 8 = x$, then $x$ is equal to

Accepted Answer

Concept & Formula: This problem is a straightforward application of the product rule of logarithms. By combining the arguments, you create a power of the base (10), allowing for easy evaluation. Product Rule: \log_a(M) + \log_a(N) = \log_a(M \cdot N) Step-by-Step Solution: Given: \log_{10}(125) + \log_{10}(8) = x Since both logarithmic terms have the same base (10), use the product rule to combine them by multiplying their arguments: \log_{10}(125 \cdot 8) = x Calculate the product inside the parenthesis: 125 \cdot 8 = 1000 Substitute this back into the equation: \log_{10}(1000) = x Now, express 1000 as a power of the base 10: 1000 = 10^3 So, the equation becomes: \log_{10}(10^3) = x Using the power rule (\log_a(a^n) = n), the logarithm of a base raised to a power evaluates to that power: 3 = x Exam Strategy & Shortcut: Recognizing common number pairs that multiply to powers of 10 is crucial for speed. 125 × 8 = 1000, 25 × 4 = 100, and 5 × 2 = 10. The moment you see 125 and 8 being added in base-10 logs, you should instantly know the argument becomes 1000. Counting the three zeros in 1000 gives you the answer 3 immediately. Common Pitfall: A common error is trying to evaluate \log_{10}(125) and \log_{10}(8) individually, which involves messy decimals or using log tables. This wastes time and introduces rounding errors. Always look to combine terms first to see if they simplify nicely. Final Answer: Therefore, the correct answer is 3.

Question 9

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 10

$(\log_5 3) 	imes (\log_3 625)$ 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 11

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

Accepted Answer

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 12

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)).

Question 13

If $\log_{10} 5 + \log_{10} (5x + 1) = \log_{10} (x + 5) + 1$, then $x$ is equal to

Accepted Answer

Concept & Formula: This equation requires condensing multiple logarithmic terms into a single term on both sides using the product rule. A key step is recognizing that an isolated integer must be converted into a logarithm before the equation can be resolved. Product Rule: \log_a(M) + \log_a(N) = \log_a(M \cdot N) Constant Conversion: 1 = \log_{10}(10) Step-by-Step Solution: Given: \log_{10}(5) + \log_{10}(5x + 1) = \log_{10}(x + 5) + 1 Step 1: Simplify the Left Side Apply the product rule to the left side, as both terms share base 10: \log_{10}(5 \cdot (5x + 1)) \log_{10}(25x + 5) Step 2: Simplify the Right Side To combine the terms on the right, we must express the integer 1 as a base-10 logarithm. We know that \log_{10}(10) = 1. Rewrite the right side: \log_{10}(x + 5) + \log_{10}(10) Now apply the product rule: \log_{10}(10 \cdot (x + 5)) \log_{10}(10x + 50) Step 3: Equate the Arguments Our simplified equation is now: \log_{10}(25x + 5) = \log_{10}(10x + 50) Since the logarithmic bases on both sides are identical, their arguments must be equal: 25x + 5 = 10x + 50 Step 4: Solve the Linear Equation Subtract 10x from both sides: 15x + 5 = 50 Subtract 5 from both sides: 15x = 45 Divide by 15: x = 3 Exam Strategy & Shortcut: Rather than solving the algebra, plug the options into the equation starting from the smallest integer. Try x = 3: Left Side: \log(5) + \log(16) = \log(80) Right Side: \log(8) + 1 = \log(8) + \log(10) = \log(80) The left and right sides match instantly. Option checking on linear log equations takes a fraction of the time compared to algebraic condensing. Common Pitfall: The most frequent mistake is ignoring the constant 1 on the right side of the equation and incorrectly equating (25x + 5) directly to (x + 5). Remember that every single term must be incorporated into the logarithm before you can drop the "log" notation from both sides. Final Answer: Therefore, the correct answer is 3.

Question 14

$\frac{1}{2}(\log x + \log y)$ will equal $\log \left(\frac{x+y}{2}\right)$ if

Accepted Answer

Concept & Formula: The problem utilizes the product rule of logarithms and basic algebraic identities to simplify logarithmic equations. \log a + \log b = \log(ab) n \log a = \log(a^n) Step-by-Step Solution: • Given: We need to simplify the left side of the equation using the product rule. (1/2)(\log x + \log y) = (1/2)\log(xy) • Calculation: Apply the power rule to move the fractional coefficient into the logarithm as an exponent. (1/2)\log(xy) = \log((xy)^{(1/2)}) = \log(\sqrt{xy}) • Equate this simplified left side to the right side of the given equation. \log(\sqrt{xy}) = \log((x+y/2)) • Since the bases of the logarithms on both sides are identical (base 10), we can drop the logarithms and equate the arguments. \sqrt{xy} = (x+y/2) • Square both sides and cross-multiply to solve for the variables algebraically. xy = ((x+y)^2/4) 4xy = x^2 + 2xy + y^2 • Rearrange the terms to form a quadratic equation. x^2 - 2xy + y^2 = 0 • Factor the perfect square trinomial. (x - y)^2 = 0 • Taking the square root of both sides gives x - y = 0, which means x = y. Exam Strategy & Shortcut: You can use the Option Verification Method here. If you substitute x = y into the original equation: Left Side: (1/2)(\log x + \log x) = (1/2)(2\log x) = \log x Right Side: \log((x+x/2)) = \log((2x/2)) = \log x Since Left Side = Right Side, option (c) is immediately verified as correct without needing to solve the full algebraic expansion. Common Pitfall: Students often incorrectly distribute the fraction or forget to square the denominator when squaring (x+y/2), leading to an incorrect polynomial and getting stuck. Always wrap the entire term in parentheses before squaring. Final Answer: Therefore, the correct answer is x = y.

Question 15

The value of $\left( \frac{1}{\log_3 60} + \frac{1}{\log_4 60} + \frac{1}{\log_5 60} \right)$ is

Accepted Answer

Concept & Formula: This problem is quickly solved by applying the Base Change Formula for logarithms, specifically the reciprocal property. When you take the reciprocal of a logarithm, the base and the argument swap places. (1/\log_a b) = \log_b a Step-by-Step Solution: • Given: We need to evaluate the sum of three reciprocal logarithms. (1/\log_3 60) + (1/\log_4 60) + (1/\log_5 60) • Calculation: Apply the reciprocal rule to each term to unify the bases. Now, all logarithms will have a common base of 60. \log_{60} 3 + \log_{60} 4 + \log_{60} 5 • Use the product rule of logarithms, which states that the sum of logs with the same base equals the log of their product (\log_b x + \log_b y = \log_b (xy)). \log_{60} (3 × 4 × 5) • Multiply the numbers inside the argument. \log_{60} (60) • Apply the identity rule (\log_a a = 1). \log_{60} 60 = 1 Exam Strategy & Shortcut: Whenever you see a sum of fractions in the format (1/\log_x N) + (1/\log_y N) ..., immediately recognize that it converts to \log_N (x × y ...). In this exact question, simply multiply the original bases 3 × 4 × 5 = 60. Since the product equals the original argument 60, the answer is instantly 1. Common Pitfall: A major mistake is attempting to convert everything to base 10 using standard change-of-base ((\log 60/\log 3)) before taking the reciprocal. While mathematically sound, it creates complex fractions that consume valuable exam time and increase calculation errors. Final Answer: Therefore, the correct answer is 1.

Question 16

The value of $(\log_3 4) (\log_4 5) (\log_5 6) (\log_6 7) (\log_7 8) (\log_8 9)$ is

Accepted Answer

Concept & Formula: This problem requires the Logarithmic Chain Rule, which is derived from the Change of Base formula. It allows you to collapse a product of logarithms when the argument of one logarithm matches the base of the next. \log_a b × \log_b c = \log_a c Step-by-Step Solution: • Given: We need to find the product of a continuous chain of logarithms. (\log_3 4) × (\log_4 5) × (\log_5 6) × (\log_6 7) × (\log_7 8) × (\log_8 9) • Calculation: Alternatively to the chain rule, express each logarithm as a fraction using the standard change of base formula to any common base (like base 10). (\log 4/\log 3) × (\log 5/\log 4) × (\log 6/\log 5) × (\log 7/\log 6) × (\log 8/\log 7) × (\log 9/\log 8) • Observe that the numerator of each fraction cancels out with the denominator of the subsequent fraction (e.g., \log 4 cancels \log 4). (\log 9/\log 3) • Rewrite this remaining fraction back into a single logarithm. \log_3 9 • Solve the final logarithm by expressing 9 as a power of the base 3. \log_3 (3^2) = 2 \log_3 3 = 2(1) = 2 Exam Strategy & Shortcut: Use the visual cancellation shortcut. In a product chain like \log_a b × \log_b c × \log_c d, just slash out the matching diagonals (the b's and c's). You are left with the very first base and the very last argument. For this question: base is 3, final argument is 9. \log_3 9 = 2. It takes 5 seconds. Common Pitfall: Students sometimes confuse logarithmic multiplication with logarithmic addition. Do not mistakenly attempt to add the arguments or bases together. Multiplication of logs only simplifies cleanly when this chain property is applicable. Final Answer: Therefore, the correct answer is 2.

Question 17

The value of $16^{\log_4 5}$ is

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Concept & Formula: This problem tests the fundamental property of exponents involving logarithms. When a base is raised to a logarithm with the same base, they cancel each other out. a^{\log_a x} = x Step-by-Step Solution: • Given: An exponential expression where the base is 16 and the exponent is a logarithm with base 4. 16^{\log_4 5} • Calculation: To apply the cancellation property, the main base (16) must match the logarithm's base (4). Rewrite 16 as a power of 4. 16 = 4^2 • Substitute this back into the original expression. (4^2)^{\log_4 5} • Apply the power of a power rule for exponents: (x^a)^b = x^{ab}. Multiply the exponents. 4^{2 × \log_4 5} • Use the logarithmic power rule (n \log_b x = \log_b (x^n)) to move the coefficient 2 inside the logarithm as an exponent for 5. 4^{\log_4 (5^2)} 4^{\log_4 25} • Now the main base (4) perfectly matches the logarithm's base (4). Apply the fundamental identity a^{\log_a x} = x. 25 Exam Strategy & Shortcut: An even faster shortcut relies on the swap property of exponential logarithms: a^{\log_b c} = c^{\log_b a}. Swap the 16 and the 5: 16^{\log_4 5} = 5^{\log_4 16} Since \log_4 16 = 2, the expression simply becomes 5^2 = 25. Common Pitfall: A frequent error is trying to apply the identity a^{\log_a x} = x prematurely without matching the bases, mistakenly answering 5 because they ignore that 16 is not 4. Always ensure bases are identical before canceling. Final Answer: Therefore, the correct answer is 25.

Question 18

If $\log x + \log y = \log (x + y)$, then

Accepted Answer

Concept & Formula: This question requires combining logarithms using the product rule, and then using basic algebraic manipulation to isolate one variable in terms of the other. \log a + \log b = \log(ab) Step-by-Step Solution: • Given: The equality mapping the sum of logs to the log of a sum. \log x + \log y = \log(x + y) • Calculation: Apply the product rule to the left side of the equation. \log(xy) = \log(x + y) • Since the logarithm function is one-to-one, and the bases are the same (base 10 implicitly), we can equate the arguments directly. xy = x + y • The options require y to be expressed in terms of x. Bring all terms containing y to one side of the equation. xy - y = x • Factor out y on the left side to isolate it. y(x - 1) = x • Divide both sides by (x - 1) to solve for y. y = (x/x - 1) Exam Strategy & Shortcut: While the algebra here is straightforward, you can also use substitution to double-check. Pick a value for x, say x = 2. xy = x + y \Rightarrow 2y = 2 + y \Rightarrow y = 2. Now test the options with x = 2: (c) y = (2-1/2) = 0.5 (Incorrect) (d) y = (2/2-1) = (2/1) = 2 (Correct) This confirms the algebraic derivation perfectly. Common Pitfall: A widespread misconception is that \log(x + y) = \log x + \log y is a general rule. It is absolutely false in standard mathematics. The problem states it as a *conditional* specific scenario to solve for a relationship, not a mathematical identity. Treat it merely as a given equation. Final Answer: Therefore, the correct answer is y = (x/x-1).

Question 19

If $\log \frac{a}{b} + \log \frac{b}{a} = \log (a + b)$, then

Accepted Answer

Concept & Formula: This problem requires the application of the product rule for logarithms to combine the terms on the left side of the equation. \log x + \log y = \log(xy) Additionally, you must recognize the fundamental logarithm identity that the log of 1 is always 0. \log 1 = 0 Step-by-Step Solution: • Given: An equation involving the sum of two logarithmic fractions. \log (a/b) + \log (b/a) = \log(a + b) • Calculation: Apply the product rule to combine the two logarithms on the left side into a single logarithm. \log ( (a/b) × (b/a) ) = \log(a + b) • Simplify the argument inside the combined logarithm. The numerator and denominator cancel each other out entirely. \log(1) = \log(a + b) • Since the logarithm of 1 to any base is always 0, the left side becomes 0. 0 = \log(a + b) • Assuming the standard base 10 (or any valid base x), convert the logarithmic equation into an exponential equation to solve for a + b. 10^0 = a + b 1 = a + b Exam Strategy & Shortcut: Notice immediately that (a/b) and (b/a) are reciprocals. When you add their logarithms, you are multiplying reciprocals inside a single log, which always yields \log(1) = 0. Therefore, you can jump straight to 0 = \log(a+b), which instantly means a+b = 1. This mental shortcut saves crucial exam time. Common Pitfall: A common mistake is incorrectly expanding \log(a+b) on the right side into \log a + \log b, which is mathematically invalid. Never confuse the product rule \log(ab) = \log a + \log b with an additive argument \log(a+b). Treat \log(a+b) as a single unbreakable entity until you convert it to exponential form. Final Answer: Therefore, the correct answer is a + b = 1.

Question 20

$\left[ \log \left( \frac{a^2}{bc} \right) + \log \left( \frac{b^2}{ac} \right) + \log \left( \frac{c^2}{ab} \right) \right]$ is equal to

Accepted Answer

Concept & Formula: This expression is simplified using the logarithmic product rule. The sum of multiple logarithms with the same base is equal to the logarithm of the product of their arguments. \log x + \log y + \log z = \log(xyz) Step-by-Step Solution: • Given: A summation of three logarithmic terms containing algebraic fractions. \log ( (a^2/bc) ) + \log ( (b^2/ac) ) + \log ( (c^2/ab) ) • Calculation: Apply the product rule to merge the three logarithms into a single logarithmic expression. Multiply the three fractional arguments together. \log ( (a^2/bc) × (b^2/ac) × (c^2/ab) ) • Combine the numerators and denominators. \log ( (a^2 \cdot b^2 \cdot c^2/(bc) \cdot (ac) \cdot (ab)) ) • Rearrange and simplify the denominator by multiplying the common variables. a \cdot a = a^2, b \cdot b = b^2, and c \cdot c = c^2. \log ( (a^2 b^2 c^2/a^2 b^2 c^2) ) • The numerator and denominator are identical, so the fraction simplifies to 1. \log(1) • The logarithm of 1, regardless of the base, is always 0. 0 Exam Strategy & Shortcut: When faced with cyclical symmetric expressions like (a^2/bc), (b^2/ac), and (c^2/ab) being added together inside logs, you should suspect they will cancel out. A quick mental check multiplying the tops (a^2b^2c^2) and bottoms (a^2b^2c^2) confirms they reduce to 1. Since \log(1) = 0, you can select option (a) in under 10 seconds without writing anything down. Common Pitfall: Some students attempt to apply the quotient rule to expand each term individually (e.g., \log(a^2) - \log(bc)). While mathematically correct, it creates a massive, sprawling expression that is highly prone to sign errors and takes significantly longer to resolve. Stick to the product rule for efficiency. Final Answer: Therefore, the correct answer is 0.

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