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 $(1, 2)$.

Using the base-change formula $\log_{a^n} b = \frac{1}{n} \log_a b$, we can rewrite the terms with a common base of 2: $\log_2 x + \log_{2^2} x + \log_{2^4} x = \frac{7}{4}$. This simplifies to $\log_2 x + \frac{1}{2}\log_2 x + \frac{1}{4}\log_2 x = \frac{7}{4}$. Factoring out $\log_2 x$, we get $(1 + \frac{1}{2} + \frac{1}{4})\log_2 x = \frac{7}{4} \Rightarrow (\frac{7}{4})\log_2 x = \frac{7}{4}$. This implies $\log_2 x = 1$, so $x = 2^1 = 2$. Option 1 (x=4) gives $\frac{7}{2}$, Option 2 (x=1) gives $0$, and Option 4 (x=16) gives $7$.

Using the properties of logarithms, specifically $\log(mn) = \log m + \log n$ and $\log(m^n) = n \log m$, we can factor 12 into its prime factors: $12 = 2^2 \times 3$. Therefore, $\log_{10} 12 = \log_{10}(2^2 \times 3) = \log_{10}(2^2) + \log_{10} 3$. Applying the power rule, this becomes $2 \log_{10} 2 + \log_{10} 3$. Substituting the given variables $a$ and $b$, we get $2a + b$. Option 1 represents $\log 6$. Option 3 represents $\log 18$. Option 4 is mathematically incorrect as $\log m^n$ is $n\log m$, not $(\log m)^n$.

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.

To find the domain of the logarithmic function $f(x) = \log_{g(x)} h(x)$, we must satisfy three critical conditions:

1. The argument must be strictly positive: $h(x) > 0 \implies x^2 - 8x + 12 > 0$. Factoring this quadratic, we get $(x-2)(x-6) > 0$, which gives $x 6$.
2. The base must be strictly positive: $g(x) > 0 \implies x - 3 > 0 \implies x > 3$.
3. The base cannot be equal to $1$: $g(x) \neq 1 \implies x - 3 \neq 1 \implies x \neq 4$.

Now, we find the intersection of these three conditions:

• From (1), $x \in (-\infty, 2) \cup (6, \infty)$.
• From (2), $x \in (3, \infty)$.
• From (3), $x \neq 4$.

The intersection of $x \in (3, \infty)$ and $x \in (-\infty, 2) \cup (6, \infty)$ is $x \in (6, \infty)$. Since all values in $(6, \infty)$ are already greater than $4$, the condition $x \neq 4$ is automatically satisfied. Thus, the domain is $(6, \infty)$.

We simplify the expression step-by-step from the innermost logarithm outward:

1. First, simplify the innermost argument: $27^3 = (3^3)^3 = 3^9$.
2. Evaluate the innermost logarithm: $\log_{3}(3^9) = 9 \log_{3}(3) = 9$.
3. Substitute this value back into the next layer: $\log_{3}(9) = \log_{3}(3^2) = 2$.
4. Substitute this value into the next layer: $\log_{2}(2) = 1$.
5. Finally, evaluate the outermost logarithm: $\log_{2}(1) = 0$.

Thus, the final value is $0$. The other options represent incomplete evaluations or algebra errors.

First, we can write the right side of the equation as a single logarithm using the property $\log_b a + \log_b c = \log_b(ac)$: $\log_5 (2x - 10) + 1 = \log_5 (2x - 10) + \log_5 5 = \log_5 [5(2x - 10)] = \log_5 (10x - 50)$.

Now, equate the arguments of the logarithms on both sides: $x^2 - 4x - 5 = 10x - 50$ $x^2 - 14x + 45 = 0$ $(x-5)(x-9) = 0 \implies x = 5 \text{ or } x = 9$.

Next, we must check these potential solutions against the domain of the original logarithmic terms:

• For $x = 5$, the term $\log_5(2x-10)$ becomes $\log_5(0)$, which is undefined because the argument of a logarithm must be strictly positive. Thus, $x = 5$ is an extraneous root.
• For $x = 9$, both arguments $x^2-4x-5 = 81-36-5 = 40 > 0$ and $2x-10 = 18-10 = 8 > 0$ are positive. Thus, $x = 9$ is the only valid solution.

To find the number of digits of a large number $N$, we calculate its common logarithm (base 10). The number of digits is given by $\lfloor \log_{10} N \rfloor + 1$ (the characteristic plus 1).

Let $N = 5^{20}$. Taking the base-10 logarithm: $\log_{10} N = \log_{10} (5^{20}) = 20 \log_{10} 5$.

We rewrite $\log_{10} 5$ using the identity $5 = \frac{10}{2}$: $\log_{10} 5 = \log_{10}\left(\frac{10}{2}\right) = \log_{10} 10 - \log_{10} 2 = 1 - 0.3010 = 0.6990$.

Now, substitute this back: $\log_{10} N = 20 \times 0.6990 = 13.98$.

The characteristic of this value is $13$. Thus, the number of digits is $13 + 1 = 14$.

To solve $\log_{0.5} (x^2 - 5x + 6) \ge -1$, we must follow two conditions:

1. Domain Condition: The argument of the logarithm must be positive: $x^2 - 5x + 6 > 0 \implies (x-2)(x-3) > 0 \implies x 3$.

2. Inequality Solving: Since the base of the logarithm ($0.5$) is less than $1$, the direction of the inequality reverses when removing the logarithm: $x^2 - 5x + 6 \le (0.5)^{-1}$ $x^2 - 5x + 6 \le 2$ $x^2 - 5x + 4 \le 0 \implies (x-1)(x-4) \le 0 \implies 1 \le x \le 4$.

Now, we find the intersection of these two sets of conditions:

• Condition 1: $x \in (-\infty, 2) \cup (3, \infty)$
• Condition 2: $x \in [1, 4]$

The intersection is $x \in [1, 2) \cup (3, 4]$. Option B is incorrect because it includes the interval $[2, 3]$ where the logarithm's argument is non-positive.

Using the change-of-base property of logarithms, we know that $\frac{1}{\log_a b} = \log_b a$.

Applying this property to each term of the series, we get: $\log_n 2 + \log_n 3 + \log_n 4 + \dots + \log_n 20$.

Using the product rule for logarithms, $\log_b x + \log_b y = \log_b(xy)$: $= \log_n (2 \cdot 3 \cdot 4 \dots 20) = \log_n (20!) = \log_n n$.

Since $\log_n n = 1$, the sum is equal to $1$.

From the second given equation: $\log_2 y = 3 \implies y = 2^3 = 8$.

Using the first equation: $\log_y x = 4 \implies x = y^4 = (2^3)^4 = 2^{12}$.

Now, evaluate the argument $\frac{x}{y^2}$: $\frac{x}{y^2} = \frac{2^{12}}{(2^3)^2} = \frac{2^{12}}{2^6} = 2^6$.

We need to find $\log_4\left(\frac{x}{y^2}\right)$: $\log_4 (2^6) = \log_{2^2} (2^6) = \frac{6}{2} \log_2 2 = 3$.

Hence, the correct option is $3$.

To solve the equation, express all logarithms with the common base $2$ using the property $\log_{b^k} a = \frac{1}{k} \log_b a$:

• $\log_4 x = \log_{2^2} x = \frac{1}{2} \log_2 x$
• $\log_{16} x = \log_{2^4} x = \frac{1}{4} \log_2 x$

Substituting these back into the equation: $\log_2 x + \frac{1}{2} \log_2 x + \frac{1}{4} \log_2 x = 7$ $\log_2 x \left(1 + \frac{1}{2} \frac{1}{4}\right) = 7$ $\log_2 x \left(\frac{7}{4}\right) = 7$ $\log_2 x = 4 \implies x = 2^4 = 16$.

Since $16 > 0$, it is a valid solution.

Since $\log a, \log b,$ and $\log c$ are in A.P., we have the standard A.P. relation: $2 \log b = \log a + \log c$.

Using the power and product properties of logarithms: $\log (b^2) = \log (ac)$.

By taking the antilog on both sides (since the log function is one-to-one): $b^2 = ac$.

This is the precise defining condition for $a, b,$ and $c$ to be in Geometric Progression (G.P.).

First, find constraints for base and argument:

1. Base $x^2 > 0$ and $x^2 \neq 1 \implies x \neq 0$ and $x \neq \pm 1$.
2. Argument $x + 2 > 0 \implies x > -2$.

Now, analyze the inequality in two cases based on the value of the base $x^2$:

Case 1: $0 < x^2 x^2 \implies x^2 - x - 2 < 0 \implies (x-2)(x+1) < 0 \implies -1 < x 1 \implies x \in (-\infty, -1) \cup (1, \infty)$. Since the base is greater than 1, the inequality sign remains the same: $x+2 0 \implies x 2$. Intersection with $x > -2$ and base constraints for Case 2 yields: $x \in (-2, -1) \cup (2, \infty)$.

Combining both cases, we get: $x \in (-2, -1) \cup (-1, 0) \cup (0, 1) \cup (2, \infty)$.

We can use the log sum property $\sum \log_b a_i = \log_b (\prod a_i)$ to simplify the sum: $\log_{10} (\tan 1^\circ \cdot \tan 2^\circ \dots \tan 89^\circ)$.

Recall the trigonometric identity $\tan(90^\circ - \theta) = \cot \theta$. This allows us to pair terms: $\tan 1^\circ \cdot \tan 89^\circ = \tan 1^\circ \cdot \cot 1^\circ = 1$. Similarly, $\tan 2^\circ \cdot \tan 88^\circ = 1$, and so on up to $\tan 44^\circ \cdot \tan 46^\circ = 1$. The middle term left unpaired is $\tan 45^\circ = 1$.

Therefore, the entire product inside the logarithm simplifies to $1$: $\log_{10}(1) = 0$.

We use the fundamental swap property of logarithms: $a^{\log_b c} = c^{\log_b a}$.

Let's apply this identity to the first term, $3^{\log_4 5}$: By swapping the base of the exponent ($3$) and the argument of the logarithm ($5$), we get: $3^{\log_4 5} = 5^{\log_4 3}$.

Thus, the expression $3^{\log_4 5} - 5^{\log_4 3}$ simplifies to: $5^{\log_4 3} - 5^{\log_4 3} = 0$.

Convert the logarithmic equation into its exponential equivalent: $x^2 - 3x = 2^2$ $x^2 - 3x = 4$ $x^2 - 3x - 4 = 0$.

Factorizing the quadratic equation: $(x - 4)(x + 1) = 0 \implies x = 4 \text{ or } x = -1$.

Now, verify both solutions in the original domain of the logarithm, which requires $x^2 - 3x > 0$:

• For $x = 4$: $4^2 - 3(4) = 16 - 12 = 4 > 0$ (Valid).
• For $x = -1$: $(-1)^2 - 3(-1) = 1 + 3 = 4 > 0$ (Valid).

Since both yield positive arguments, there are exactly $2$ valid real solutions.

Using the change-of-base formula, we write $\log_a b = \frac{\log b}{\log a}$: $\frac{\log 2}{\log 3} \cdot \frac{\log 3}{\log 4} \cdot \frac{\log 4}{\log 5} \dots \frac{\log (n-1)}{\log n} = \frac{1}{10}$.

Notice that the terms cancel out telescopically:

• The denominator $\log 3$ cancels with the next numerator $\log 3$.
• The denominator $\log 4$ cancels with the next numerator $\log 4$, and so on.

After all cancellations, only the first numerator and the last denominator remain: $\frac{\log 2}{\log n} = \frac{1}{10} \implies \log_n 2 = \frac{1}{10}$.

In exponential form, this is: $n^{1/10} = 2 \implies n = 2^{10} = 1024$.

First, apply the change-of-base reciprocal rule $\frac{1}{\log_x y} = \log_y x$ to rewrite the terms: $\log_a (c+b) + \log_a (c-b)$.

Using the product property of logarithms: $= \log_a [(c+b)(c-b)] = \log_a (c^2 - b^2)$.

We are given the Pythagorean identity relation $a^2 + b^2 = c^2 \implies c^2 - b^2 = a^2$.

Substitute $a^2$ into the expression: $= \log_a (a^2) = 2 \log_a a = 2$.