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WAEC Further Mathematics 2010 Paper

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Question : 4

Total: 40

If

16^{3x} = \; \frac{1}{4} (32^{x - 1})

, find the value of x.

$-1$
$\; \frac{-1}{3}$
$\; \frac{-3}{7}$
$\; \frac{-5}{19}$

Solution:

Concept:

Rewrite both sides as powers of the same base, then compare exponents.

Explanation:

Given:

16^{3x} = \frac{1}{4}(32^{x-1})

.

Express each base as a power of

2

.

16 = 2^4

and

32 = 2^5

, and

\frac{1}{4} = 2^{-2}

.

So,

(2^4)^{3x} = 2^{-2} \cdot (2^5)^{x-1}

.

Using index laws:

2^{12x} = 2^{-2+5x-5}

.

Therefore,

12x = 5x - 7

.

Subtract

5x

from both sides:

7x = -7

.

Hence,

x = -1

.

Answer:

x = -1

Option A.

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