Questions & Answers: "Logarithm"

To define the logarithm $\log_b a$, three conditions must be met: 1) The argument $a > 0$, 2) The base $b > 0$, and 3) The base $b \neq 1$. For $f(x) = \log_{(x-1)}(5-x)$, we require: (i) $5-x > 0 \Rightarrow x 0 \Rightarrow x > 1$. (iii) $x-1 \neq 1 \Rightarrow x \neq 2$. Combining these, $x$ must be in the interval $(1, 5)$ but cannot be $2$. Thus, the domain is $(1, 2) \cup (2, 5)$. Option 1 is wrong because it includes $x=2$, which makes the base 1. Option 3 is wrong because logs are undefined for zero or negative numbers. Option 4 is wrong because it unnecessarily excludes the interval $(1, 2)$.

To solve log₂(x) + log₂(x - 2) = 3, we use the product rule of logarithms: logₐ(m) + logₐ(n) = logₐ(m*n). This gives log₂(x(x - 2)) = 3. Converting this to exponential form, we get x(x - 2) = 2³, which simplifies to x² - 2x = 8, or x² - 2x - 8 = 0. Factoring the quadratic equation gives (x - 4)(x + 2) = 0, resulting in x = 4 or x = -2. However, the definition of a logarithm logₐ(y) requires y > 0. For log₂(x) to be defined, x > 0, and for log₂(x - 2) to be defined, x > 2. Therefore, x = -2 is an extraneous solution and must be rejected. The only valid solution is x = 4.

This question utilizes the base change property: 1/logₐ(b) = log_b(a). Applying this to each term, the expression becomes log₃₀(3) + log₃₀(2) + log₃₀(5). Using the product rule of logarithms (logₐ(m) + logₐ(n) + logₐ(p) = logₐ(mnp)), we combine the terms: log₃₀(3 * 2 * 5) = log₃₀(30). Since any logarithm where the base and the argument are the same (logₐ(a)) equals 1, the final value is 1. Options 0 and 30 are mathematically inconsistent with these logarithmic identities.

To find the value, let log_{2\sqrt{3}} (1728) = x. This implies (2\sqrt{3})^x = 1728. Squaring both sides or expressing 1728 in terms of prime factors: 1728 = 12^3. Now, (2\sqrt{3})^2 = 4 * 3 = 12. Therefore, 1728 = ( (2\sqrt{3})^2 )^3 = (2\sqrt{3})^6. Comparing the powers, x = 6. The other options are incorrect because they do not satisfy the exponential form of the logarithmic equation.

Using the quotient rule of logarithms, log(a/b) = log a - log b. log_{10} 1.5 = log_{10} (3/2) = log_{10} 3 - log_{10} 2. Substituting the given values: 0.4771 - 0.3010 = 0.1761. Option A is the sum (product rule), which is incorrect. Other options are mathematical errors.

The fundamental property of logarithms states that a^{log_a n} = n. Applying this to both terms: 7^{log_7 5} = 5 and 2^{log_2 3} = 3. Adding them gives 5 + 3 = 8. The other options result from incorrectly multiplying the numbers or ignoring the log properties.

To find the number of digits in N = 2^{50}, take the common log: log N = log(2^{50}) = 50 * log 2 = 50 * 0.3010 = 15.05. The characteristic is 15. The number of digits in a number is (characteristic + 1). Therefore, digits = 15 + 1 = 16. Option A is just the characteristic.

This is the Change of Base Formula. log_b a = (log_c a) / (log_c b) for any positive base c ≠ 1. Option D is the reciprocal (1 / log_b a), and options B and C are common misconceptions about log properties that do not exist.

When the base of a logarithm is between 0 and 1 (decreasing function), removing the logs reverses the inequality sign. So, log_{0.5} (x - 1) > log_{0.5} 4 implies x - 1 < 4, or x 0, so x > 1. Combining these gives 1 < x < 5. Option A incorrectly keeps the inequality direction.

If a, b, c are in G.P., then b^2 = ac. Taking the log of both sides: log(b^2) = log(ac). Using properties: 2 log b = log a + log c. This is the condition for log a, log b, log c to be in Arithmetic Progression (A.P.), where the middle term is the average of the outer terms.

Starting from the outside: log_2(...) = 1 means log_3 (log_2 x) = 2^1 = 2. Next, log_3(...) = 2 means log_2 x = 3^2 = 9. Finally, log_2 x = 9 means x = 2^9 = 512. (Note: In this specific formatting, both 2^9 and 512 represent the same value, but for curriculum accuracy, 512 is the simplified integer solution).

To find the number of digits in a number N, we find log₁₀ N. Here, N = 6²⁰. log₁₀(6²⁰) = 20 * log₁₀(2 * 3) = 20 * (log₁₀ 2 + log₁₀ 3) = 20 * (0.3010 + 0.4771) = 20 * 0.7781 = 15.562. The characteristic is 15. The number of digits is equal to characteristic + 1, which is 15 + 1 = 16. Option B is correct.

Using the product rule log a + log b = log(ab), we get log₃((x + 1)(x - 1)) = log₃ 8. This implies (x + 1)(x - 1) = 8, so x² - 1 = 8, which means x² = 9. This gives x = 3 or x = -3. However, the domain of a logarithm requires the argument to be positive. If x = -3, (x+1) and (x-1) are negative, making the original logs undefined. Thus, only x = 3 is valid.

First, evaluate log₅ 125. Since 125 = 5³, log₅ 5³ = 3. Second, evaluate log₁₀ 0.001. Since 0.001 = 10⁻³, log₁₀ 10⁻³ = -3. Multiplying the two results: 3 * (-3) = -9. Therefore, option B is the correct answer.

For log(g(x)) to be defined, g(x) must be greater than zero. So, x² - 5x + 6 > 0. Factoring the quadratic gives (x - 2)(x - 3) > 0. Using the wavy curve method or testing intervals: the expression is positive when x 3. Thus, the domain is (-∞, 2) ∪ (3, ∞). Option A is wrong because it's the interval where the quadratic is negative.

According to the fundamental identity of logarithms, a^(logₐ x) = x for any x > 0 and base a > 0, a ≠ 1. Applying this here, 7^(log₇ 11) = 11. This is a direct application of the definition of a logarithm as the inverse of exponentiation.

Converting the logarithmic form logₓ 256 = 8 into exponential form, we get x⁸ = 256. We know that 256 = 2⁸. Therefore, x⁸ = 2⁸. Since the exponents are equal, the bases must be equal, so x = 2. Option A is the correct answer.

Using the change of base formula log_a b = (log b / log a), the expression becomes: (log 3 / log 2) * (log 4 / log 3) * (log 5 / log 4) * (log 8 / log 5). Most terms cancel out, leaving (log 8 / log 2). This simplifies to log₂ 8. Since 8 = 2³, log₂ 2³ = 3. Thus, option B is correct.

When the base of a logarithm is between 0 and 1 (as is 0.5), the inequality sign flips when comparing the arguments. So, log₀.₅ (x - 3) 5. This leads to x > 8. Also, for the logarithm to be defined, x - 3 > 0 (x > 3), which is already satisfied by x > 8. Therefore, the solution is x > 8.

Using change of base: log₅ 12 = log₁₀ 12 / log₁₀ 5. Numerator: log₁₀ 12 = log₁₀(2² * 3) = 2log₁₀ 2 + log₁₀ 3 = 2a + b. Denominator: log₁₀ 5 = log₁₀(10/2) = log₁₀ 10 - log₁₀ 2 = 1 - a. Thus, log₅ 12 = (2a + b) / (1 - a). Option A is correct.

Using the property 1 / log_a b = log_b a, the expression transforms to: log_n 2 + log_n 3 + log_n 4 + ... + log_n 10. By the product rule log x + log y = log(xy), this becomes log_n (2 * 3 * 4 * ... * 10). Since 1 * 2 * ... * 10 is 10!, the expression equals log_n (10!). Option A is correct.