Test Index

UPSC NDA I 2016 Mathematics Solved Paper

© examsnet.com

Question : 118 of 120

Marks: +1, -0

If

\log_{a}(ab) = x

\log_{b}(ab)

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

Solution:

Concept:

Use logarithm properties:

\log_a(MN) = \log_a M + \log_a N

,

\log_a a = 1

, and

\log_a b = \frac{1}{\log_b a}

.

Explanation:

Given:

\log_a(ab) = x

.

We need

\log_b(ab)

.

First, find

\log_a b

from the given equation:

\log_a(ab) = \log_a a + \log_a b = 1 + \log_a b = x

So

\log_a b = x - 1

.

Now express

\log_b(ab)

in terms of

\log_a b

:

\log_b(ab) = \log_b a + \log_b b = \log_b a + 1

Since

\log_b a = \frac{1}{\log_a b}

, we have:

\log_b(ab) = \frac{1}{\log_a b} + 1 = \frac{1}{x-1} + 1

Combine terms:

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

.

Thus,

\log_b(ab) = \frac{x}{x-1}

.

This matches option D.

Answer:

Option D:

\frac{x}{x-1}

© examsnet.com

Go to Question:

Prev Question

Next Question