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WAEC Mathematics 2015 Paper

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Question : 7

Total: 45

If 2ⁿ = y, Find 2

^{\left(2 + \frac{n}{3}\right)}

4y $^{\frac{1}{3}}$
4y $^{-3}$
2y $^{\frac{1}{3}}$
2y $^{-3}$

Solution:

Concept:

Use laws of exponents:

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

and

\left(a^m\right)^n = a^{mn}

.

Explanation:

Given

2^n = y

.

We need

2^{\left(2 + \frac{n}{3}\right)}

.

Split the exponent:

2^{\left(2 + \frac{n}{3}\right)} = 2^2 \times 2^{\frac{n}{3}}

.

Since

2^2 = 4

, we get

4 \times 2^{\frac{n}{3}}

.

Now write

2^{\frac{n}{3}}

as

\left(2^n\right)^{\frac{1}{3}}

.

Using

2^n = y

, this becomes

y^{\frac{1}{3}}

.

Therefore,

2^{\left(2 + \frac{n}{3}\right)} = 4y^{\frac{1}{3}}

.

Answer:

4y^{\frac{1}{3}}

Option A.

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