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Question : 3 of 86

Marks: +1, -0

If,

\left(\frac{a}{b}\right)^{x-1}=\left(\frac{b}{a}\right)^{x-3}

then the value of

x

is

$\frac{1}{2}$
1
2
$\frac{7}{2}$

Solution:

Concept:

Rewrite

\frac{b}{a}

as

\left(\frac{a}{b}\right)^{-1}

and equate exponents.

Explanation:

Given

\left(\frac{a}{b}\right)^{x-1} = \left(\frac{b}{a}\right)^{x-3}

.

Since

\frac{b}{a} = \left(\frac{a}{b}\right)^{-1}

, the equation becomes:

\left(\frac{a}{b}\right)^{x-1} = \left(\frac{a}{b}\right)^{-(x-3)}.

Equating exponents (as base

\frac{a}{b} \neq 0, \pm 1

in general):

x-1 = -(x-3)

x-1 = -x+3

2x = 4

x = 2

.

Answer:

x = 2

, which corresponds to option C.

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