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CTET Paper II (Mathematics and Science) Sep 2016 Solved Paper

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Question : 43 of 150

Marks: +1, -0

The value of m such that

(\frac{2}{9})^{3}\times (\frac{4}{81})^{-6}=(\frac{2}{9})^{2m-1}

is

$-4$
3
$-6$
0

Solution:

Concept:

Use the property

(a^m)^n = a^{mn}

and

a^m \times a^n = a^{m+n}

to equate exponents when bases are the same.

Explanation:

First, rewrite

\frac{4}{81}

as

(\frac{2}{9})^2

.

Then

(\frac{4}{81})^{-6} = [(\frac{2}{9})^2]^{-6} = (\frac{2}{9})^{-12}

.

Multiply:

(\frac{2}{9})^3 \times (\frac{2}{9})^{-12} = (\frac{2}{9})^{3 + (-12)} = (\frac{2}{9})^{-9}

.

Equate to the right side:

(\frac{2}{9})^{-9} = (\frac{2}{9})^{2m-1}

.

Since bases are equal, set exponents equal:

-9 = 2m - 1

.

Solve:

2m = -9 + 1 = -8

, so

m = -4

.

Answer:

m = -4

(Option A)

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