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:

Concept:

Express both sides of each equation using the same base, then compare exponents to form linear equations.

Explanation:

Write

16

and

\frac{1}{32}

as powers of

2

16 = 2^4

and

\frac{1}{32} = 2^{-5}

From

2^{x+y} = 16

, we get

2^{x+y} = 2^4

, so

x + y = 4

From

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

, rewrite

4

2^2

4^{x-y} = (2^2)^{x-y} = 2^{2x-2y}

Thus

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

, giving

2x - 2y = -5

Divide the second equation by

2

x - y = -\frac{5}{2}

Solve simultaneously:

x + y = 4

and

x - y = -\frac{5}{2}

Add the equations:

2x = 4 - \frac{5}{2} = \frac{3}{2}

, so

x = \frac{3}{4}

Substitute

x = \frac{3}{4}

into

x + y = 4

y = 4 - \frac{3}{4} = \frac{13}{4}

Answer:

x = \frac{3}{4}

and

y = \frac{13}{4}

, which is Option B.

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