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WAEC Further Mathematics 2010 Paper

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Question : 4

Total: 40

If

16^{3x} = \; \frac{1}{4} (32^{x - 1})

, find the value of x.

$-1$
$\; \frac{-1}{3}$
$\; \frac{-3}{7}$
$\; \frac{-5}{19}$

Solution:

$16^{3x} = \; \frac{1}{4} (32^{x - 1})$

$(2^{4})^{3x} = (2^{-2})((2^{5})^{x - 1})$

$2^{12x} = 2^{-2 + 5x - 5}$

$12x = -7 + 5x$

$7x = -7 \implies x = -1$

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