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WAEC Further Mathematics 2025 Paper

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Question : 7

Total: 40

Given that

\frac{1}{8^{2-3y}}

^{y + 2}

, find y.

1
$\frac{1}{5}$
$\frac{7}{8}$
$\frac{1 1}{5}$

Solution:

Concept:

Write both sides with the same base and then equate the exponents.

Explanation:

We have:

\frac{1}{8^{2-3y}} = 2^{y+2}

Express

\frac{1}{8^{2-3y}}

as

8^{-(2-3y)}

.

Since

8 = 2^3

, rewrite

8^{-(2-3y)}

as

2^{3(-(2-3y))}

.

This simplifies to

2^{-6+9y}

.

Now the equation is

2^{-6+9y} = 2^{y+2}

.

Equate the exponents:

-6+9y = y+2

.

Simplify:

9y - y = 2 + 6

.

So

8y = 8

, giving

y = 1

.

Answer:

y = 1

(Option A).

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