ABC26GN2051 · Logarithms
Subject: General Aptitude · Chapter: Logarithms · Exam: 2002 · Marks: · Difficulty:
If $\log _{a}(a b)=x$, then $\log _{b}(a b)$ is
(a)$\frac{1}{x}$
(b)$\frac{x}{x+1}$
(c)$\frac{x}{1-x}$
(d)$\frac{x}{x-1}$
Answer
Explanation
$\log _{a}(a b)=x \Leftrightarrow \frac{\log a b}{\log a}=x \Leftrightarrow \frac{\log a+\log b}{\log a}=x$ $$\begin{aligned} & \Leftrightarrow 1+\frac{\log b}{\log a}=x \Leftrightarrow \frac{\log b}{\log a}=x-1 \Leftrightarrow \frac{\log a}{\log b}=\frac{1}{x-1} \\ & \Leftrightarrow 1+\frac{\log a}{\log b}=1+\frac{1}{x-1} \Leftrightarrow \frac{\log b}{\log b}+\frac{\log a}{\log b}=\frac{x}{x-1} \\ & \Leftrightarrow \frac{\log b+\log a}{\log b}=\frac{x}{x-1} \Leftrightarrow \frac{\log (a b)}{\log b}=\frac{x}{x-1} \\ & \Leftrightarrow \log _{b}(a b)=\frac{x}{x-1} . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
Open in whiteboard · Browse this chapter in the app