If $${\log _a}\left( {ab} \right) = x{\text{,}}\,$$ then $${\log _b}\left( {ab} \right)$$ is -
A. $$\frac{1}{x}$$
B. $$\frac{x}{{x + 1}}$$
C. $$\frac{x}{{1 - x}}$$
D. $$\frac{x}{{x - 1}}$$
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {\text{lo}}{{\text{g}}_a}\left( {ab} \right) = x \cr & \Rightarrow \frac{{\log ab}}{{\log a}} = x \cr & \Rightarrow \frac{{\log a + \log b}}{{\log a}} = x \cr & \Rightarrow 1 + \frac{{\log b}}{{\log a}} = x \cr & \Rightarrow \frac{{\log b}}{{\log a}} = x - 1 \cr & \Rightarrow \frac{{\log a}}{{\log b}} = \frac{1}{{x - 1}} \cr & \Rightarrow 1 + \frac{{\log a}}{{\log b}} = 1 + \frac{1}{{x - 1}} \cr & \Rightarrow \frac{{\log b}}{{\log b}} + \frac{{\log a}}{{\log b}} = \frac{x}{{x - 1}} \cr & \Rightarrow \frac{{\log b + \log a}}{{\log b}} = \frac{x}{{x - 1}} \cr & \Rightarrow \frac{{{\text{log}}\left( {ab} \right)}}{{\log b}} = \frac{x}{{x - 1}} \cr & \Rightarrow {\text{lo}}{{\text{g}}_b}\left( {ab} \right) = \frac{x}{{x - 1}} \cr} $$Related Questions on Logarithm
Which of the following statements is not correct?
A. log10 10 = 1
B. log (2 + 3) = log (2 x 3)
C. log10 1 = 0
D. log (1 + 2 + 3) = log 1 + log 2 + log 3
$${{\log \sqrt 8 } \over {\log 8}}$$ is equal to:
A. $$\frac{1}{6}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$\frac{1}{8}$$
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