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

Assessment: JAMB UTME - Mathematics - 2019 Subject: General Mathematics

Question 1 Report

x =

\frac{3}{4}

, y =

\frac{11}{4}

x =

\frac{3}{4}

, y =

\frac{13}{4}

x =

\frac{2}{3}

, y =

\frac{4}{5}

x =

\frac{2}{3}

, y =

\frac{13}{4}

Answer Details

^{x + y}

= 16 ; 4

^{x - y}

\frac{1}{32}

⟹ 2^{x + y} = 2^{4}

x + y = 4 . . . (1)

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

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

⟹ 2 x - 2 y = - 5 . . . (2)

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

\frac{3}{4}

and y =

\frac{13}{4}

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