We use the product property of logarithms, log(a) + log(b) + log(c) = log(abc). Multiplying the three arguments gives (x)/(y) × (y)/(z) × (z)/(x) = 1. Since log(1) = 0, the sum evaluates directly to 0.

We use the product property of logarithms, log(a) + log(b) + log(c) = log(abc). Multiplying the three arguments gives (x)/(y) × (y)/(z) × (z)/(x) = 1. Since log(1) = 0, the sum evaluates directly to 0.

Multiplying the arguments gives 6 × 12 × 8 = 576. Since 576 = 24², we simplify log_24(24²) to 2 × 1 = 2, proving the given statement.

We rewrite 1 as log_c c. Adding x = log_c(ab) yields 1 + x = log_c(abc). Taking the reciprocal gives (1)/(1 + x) = log_abc c. By identical symmetry, (1)/(1 + y) = log_abc a and (1)/(1 + z) = log_abc b.

We rewrite 1 as log_c c. Adding x = log_c(ab) yields 1 + x = log_c(abc). Taking the reciprocal gives (1)/(1 + x) = log_abc c. By identical symmetry, (1)/(1 + y) = log_abc a and (1)/(1 + z) = log_abc b.

Using the power rule k log(u) = log(u^k), we rewrite the terms as log((36)/(49)), log((9)/(4)), and log((196)/(27)). Then, we combine them into a single logarithm using the product rule.

Cancelling common factors: (196)/(49) = 4, (36)/(4) = 9, and (9 × 4)/(27) = (36)/(27) = (4)/(3). Multiplying 9 × (1)/(3) × 4 = 12. Therefore, the expression simplifies to log 12.

1. Show that: $\log \frac{x}{y} + \log \frac{y}{z} + \log \frac{z}{x} = 0$ 2. Show that: $\frac{1}{\log_6 24} + \frac{1}{\log_{12} 24} + \frac{1}{\log_8 24} = 2$ 3. If $x = \log_c ab$ , $y = \log_a bc$ , $z = \log_b ca$ , prove that: $\frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} = 1$ 4. Prove that: $2\log\frac{6}{7} + \frac{1}{2}\log\frac{81}{16} + \log\frac{196}{27} = \log 12$

Answer: All four logarithmic statements are proven: 1. $\log\left(\frac{x}{y}\right) + \log\left(\frac{y}{z}\right) + \log\left(\frac{z}{x}\right) = 0$ 2. $\frac{1}{\log_6 24} + \frac{1}{\log_{12} 24} + \frac{1}{\log_8 24} = 2$ 3. $\frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} = 1$ 4. $2\log\left(\frac{6}{7}\right) + \frac{1}{2}\log\left(\frac{81}{16}\right) + \log\left(\frac{196}{27}\right) = \log 12$

Step-by-step solution

Step 1: Solve Question 1 using Logarithmic Product Rule

We use the product property of logarithms, $\log(a) + \log(b) + \log(c) = \log\text{(abc)}$ . Multiplying the three arguments gives $\frac{x}{y} \times \frac{y}{z} \times \frac{z}{x} = 1$ . Since $\log(1) = 0$ , the sum evaluates directly to $0$ .

Step 2: Solve Question 2 using Base-Change Property

Applying the reciprocal base property $\frac{1}{\log_b a} = \log_a b$ , the expression becomes $\log_{24} 6 + \log_{24} 12 + \log_{24} 8$ . Now all logarithms share the common base $24$ , so we combine them by multiplying their arguments.

Step 3: Evaluate the Product for Question 2

Multiplying the arguments gives $6 \times 12 \times 8 = 576$ . Since $576 = 24^2$ , we simplify $\log_{24}(24^2)$ to $2 \times 1 = 2$ , proving the given statement.

Step 4: Solve Question 3: Simplify Denominators

We rewrite $1$ as $\log_c c$ . Adding $x = \log_c(ab)$ yields $1 + x = \log_c\text{(abc)}$ . Taking the reciprocal gives $\frac{1}{1 + x} = \log_{abc} c$ . By identical symmetry, $\frac{1}{1 + y} = \log_{abc} a$ and $\frac{1}{1 + z} = \log_{abc} b$ .

Step 5: Sum the Symmetrical Terms for Question 3

We add the three simplified fractions together: $\log_{abc} c + \log_{abc} a + \log_{abc} b$ . Using the product rule, this equals $\log_{abc}\text{(abc)}$ , which equals $1$ .

Step 6: Solve Question 4: Apply Power Rule and Combine

Using the power rule $k \log(u) = \log(u^k)$ , we rewrite the terms as $\log\left(\frac{36}{49}\right)$ , $\log\left(\frac{9}{4}\right)$ , and $\log\left(\frac{196}{27}\right)$ . Then, we combine them into a single logarithm using the product rule.

Step 7: Simplify Fractions for Question 4

Cancelling common factors: $\frac{196}{49} = 4$ , $\frac{36}{4} = 9$ , and $\frac{9 \times 4}{27} = \frac{36}{27} = \frac{4}{3}$ . Multiplying $9 \times \frac{1}{3} \times 4 = 12$ . Therefore, the expression simplifies to $\log 12$ .

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