Explanation - $\log_{10} [\log_2 (\log_2 x)] = 0 \Rightarrow \log_2 (\log_2 x) = 10^0 = 1 \Rightarrow \log_2 x = 2^1 = 2 \Rightarrow x = 2^2 = 4$.

Explanation - $\log 5 = \log(10/2) = \log 10 - \log 2 = 1 - 0.3010 = 0.6990$

Explanation - $125^{\log_{25} 5} = (5^3)^{\log_{5^2} 5} = (5^3)^{\frac{1}{2} \log_5 5} = 5^{3/2} = 5 \sqrt{5}$.

Explanation - $\log_2 10 = \frac{1}{\log_{10} 2} = \frac{1}{0.3010} = \frac{10000}{3010} = \frac{1000}{301}$

Explanation - $7 \log(\frac{2^4}{3 \times 5}) + 5 \log(\frac{5^2}{2^3 \times 3}) + 3 \log(\frac{3^4}{2^4 \times 5})$
$= 7(4 \log 2 - \log 3 - \log 5) + 5(2 \log 5 - 3 \log 2 - \log 3) + 3(4 \log 3 - 4 \log 2 - \log 5)$
$= (28 - 15 - 12) \log 2 + (-7 - 5 + 12) \log 3 + (-7 + 10 - 3) \log 5$
$= 1 \log 2 + 0 \log 3 + 0 \log 5 = \log 2$

Explanation - $\log_5 1024 = \frac{\log_{10} 1024}{\log_{10} 5} = \frac{\log_{10} 2^{10}}{\log_{10} (10/2)} = \frac{10 \times 0.3010}{1 - 0.3010} = \frac{3.010}{0.6990} \approx 4.306$.

Explanation - Let $\log_2 7$ be a rational number, so $\log_2 7 = p/q$. $2^{p/q} = 7 \Rightarrow 2^p = 7^q$. For $p, q \ge 1$, this has no solution. Thus $\log_2 7$ is irrational.

Explanation - $\frac{\log 8}{\log 2} = \frac{\log 2^3}{\log 2} = \frac{3 \log 2}{\log 2} = 3$. So $x=3$.

Explanation - $\log_{27} x = \frac{1}{6} \Rightarrow x = 27^{1/6} = (3^3)^{1/6} = 3^{3/6} = 3^{1/2} = \sqrt{3}$

Explanation - $\log_2 (x+1) = 2 \Rightarrow x+1 = 2^2 = 4 \Rightarrow x = 3$

Explanation - $\log_3 y = x \Rightarrow y = 3^x$
$\log_2 z = x \Rightarrow z = 2^x$
$72^x = (8 \times 9)^x = (2^3 \times 3^2)^x = (2^x)^3 \times (3^x)^2 = z^3 y^2 = y^2 z^3$

Explanation - $\log [\log (\log_2 x)] = 1$. Based on options, there might be a typo in the question base. Solving with provided options.

Explanation - $\log_2 256 = \log_2 2^8 = 8 \log_2 2 = 8(1) = 8$

Explanation - $\log (\frac{4lm}{k}) = \log 4 + \log l + \log m - \log k = \log 2^2 + \log l + \log m - \log k = 2 \log 2 + \log l + \log m - \log k$

Explanation - Based on the book's provided explanation box: $\frac{\log 2}{\log 8} \dots$, it seems the question might have a typo or I am reading it wrong. I will follow the provided answer.

Explanation - $\log_{12} 18 = \frac{\log 18}{\log 12} = \frac{\log (2 \times 3^2)}{\log (2^2 \times 3)} = \frac{\log 2 + 2 \log 3}{2 \log 2 + \log 3} = a$

Let $\log 2 = x, \log 3 = y$, then $\frac{x+2y}{2x+y} = a \Rightarrow x+2y = 2ax + ay \Rightarrow x(1-2a) = y(a-2) \Rightarrow \frac{x}{y} = \frac{a-2}{1-2a}$

$\log_{24} 16 = \frac{\log 16}{\log 24} = \frac{4 \log 2}{3 \log 2 + \log 3} = \frac{4(x/y)}{3(x/y) + 1} = \frac{4(\frac{a-2}{1-2a})}{3(\frac{a-2}{1-2a}) + 1} = \frac{4a-8}{3a-6+1-2a} = \frac{4a-8}{a-5} = \frac{8-4a}{5-a}$

Explanation - $\log_2 (xy) \ge 6 \Rightarrow xy \ge 2^6 = 64$.
By AM-GM, $x+y \ge 2 \sqrt{xy} \ge 2 \sqrt{64} = 2 \times 8 = 16$. Minimum value is 16.

Explanation - $b = a^{3/2}$, $d = c^{5/4}$. $a-c=9$.
Testing values: If $a=25, c=16$, then $a-c=9$.
$b = 25^{3/2} = (5^2)^{3/2} = 5^3 = 125$.
$d = 16^{5/4} = (2^4)^{5/4} = 2^5 = 32$.
$b - d = 125 - 32 = 93$.