If 2^{x+y} = 16 and 4^{x-y} = 1/32, find the values of x and y.

Correct Answer is : x = 343/443, y = 13413/4413

Correct Answer is : x = 343/443, y = 13413/4413

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JAMB Mathematics 2019 Paper

Question : 40

Total: 73

If 2

^{x+y}

^{x-y} = \frac{1}{32}

, find the values of x and y.

Solution:

2 $^{x+y}$ = 16 ; 4 $^{x-y}$ = $\frac{1}{32}$ .

$\implies 2^{x+y} = 2^4$

$x + y = 4 \dots (1)$

$2^{2(x - y)} = 2^{-5}$

$2^{2x - 2y} = 2^{-5}$

$\implies 2x - 2y = -5 \dots (2)$

Solving the equations (1) and (2) simultaneously, we have

x = $\frac{3}{4}$ and y = $\frac{13}{4}$

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