$Which of the following is correct?$

Moderate

A

$If a^{2} + 4 b^{2} = 12 a b, then$ $\log (a + 2 b) = \frac{1}{2} (\log a + \log b)$
B

$If \frac{\log x}{b - c} = \frac{\log y}{c - a} = \frac{\log z}{a - b}$ $then x^{a} \cdot y^{b} \cdot z^{c} = a b c$
C

$\frac{1}{\log_{xy} xy z} + \frac{1}{\log_{yz} xy z} + \frac{1}{\log_{zx} xyz} = 2$
D

All are correct

Solution

$\begin{matrix} (1) \log (a + 2 b) = \frac{1}{2} \log {(a + 2 b)}^{2} \\ = \frac{1}{2} \log (a^{2} + 4 b^{2} + 4 a b) \\ = \frac{1}{2} \log (12 a b + 4 a b) \\ = \frac{1}{2} \log (2^{4} \cdot ab) \\ = \frac{1}{2} (4 \log 2 + \log a + \log b) \end{matrix}$
$\begin{matrix} (2) Let \frac{\log x}{b - c} = \frac{\log y}{c - a} = \frac{\log z}{a - b} = k \\ \Rightarrow \log x = k (b - c), \log y = k (c - a) \\ \log z = k (a - b) \\ x^{a} \cdot y^{b} \cdot z^{c} = p^{k [a (b - c)) + b (c - a) + c (a - b)]} \\ {=p}^{k (0)} = 1 \end{matrix}$
$\begin{matrix} (3) Given expression is = \log_{x y z} x y + \log_{x y z} y z + \log_{x y z} z x \\ = \log_{x y z} (x y \cdot y z \cdot z x) = \log_{x y z} (x^{2} \cdot y^{2} \cdot z^{2}) \\ = 2 \log_{xyz} (xyz) = 2 \times 1 = 2 \end{matrix}$

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