JAMB 1986 · UME · Q25

If $5^{x + 2y} = 5$ and $4^{x + 3y} = 16$ , find $3^{x + y}$ .

Worked solution (try it first)

$5^{x + 2y} = 5^1$ , so $x + 2y = 1$ .
$4^{x + 3y} = 16 = 4^2$ , so $x + 3y = 2$ .
Subtract the first equation from the second: $y = 1$ .
Then $x = 1 - 2 = -1$ .
So $x + y = 0$ and $3^{x + y} = 3^0 = 1$ , option B.

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