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WAEC Further Mathematics 2014 Paper

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Question : 6

Total: 40

If

8^{x} \div \left( \frac{1}{4} \right)^{y} = 1

\log_{2}(x - 2y) = 1

, find the value of (x - y).

$\frac{5}{4}$
$\frac{3}{5}$
$1$
$\frac{2}{3}$

Solution:

Concept:

Express all terms as powers of

2

, then solve the two resulting linear equations.

Explanation:

Since

8=2^3

and

\frac14=2^{-2}

, rewrite the first condition:

8^x \div \left(\frac14\right)^y = 1

becomes

(2^3)^x \div (2^{-2})^y = 1

.

So

2^{3x} \div 2^{-2y}=2^0

, which gives

2^{3x+2y}=2^0

.

Equating exponents:

3x+2y=0

. ...(1)

Now use

\log_2(x-2y)=1

.

Convert to exponential form:

x-2y=2^1=2

. ...(2)

Add equations (1) and (2):

(3x+2y)+(x-2y)=0+2

, so

4x=2

.

Hence

x=\frac12

.

Substitute

x=\frac12

into

x-2y=2

:

\frac12-2y=2

, so

-2y=\frac32

, giving

y=-\frac34

.

Therefore,

x-y=\frac12-\left(-\frac34\right)=\frac12+\frac34=\frac54

.

Answer:

(x-y)=\frac54

, which is Option A.

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