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Definite Integral as Limit of Sum

If logaab =...

Question

If $l o g_{a} (a b) = x$ , then $l o g_{b} (a b)$ is

A

\frac{x}{(1 + x)}

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B

\frac{1}{x}

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C

\frac{x + 1}{x - 1}

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D

\frac{x}{x - 1}

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Solution

The correct option is C $\frac{x}{x - 1}$
We have,
$l o g_{a} (a b) = x$
$l o g_{a} a + l o g_{a} b = x$ $[l o g_{a} m n = l o g_{a} m + l o g_{a} n]$
$1 + l o g_{a} b = x$ $[l o g_{a} a = 1]$
$l o g_{a} b = x - 1$ $. . . . . . (1)$

Since,
$= l o g_{b} (a b)$
$= l o g_{b} a + l o g_{b} b$
$= l o g_{b} a + 1$
$= \frac{1}{l o g_{a} b} + 1$ $[\frac{1}{l o g_{n} m} = l o g_{m} n]$
$= \frac{1}{x - 1} + 1$
$= \frac{1 + x - 1}{x - 1}$
$= \frac{x}{x - 1}$

Hence, this is the answer.

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