Solve the following equation:
${log}_{2} (9 - 2^{x}) = 3 - x$

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Solution

$l o g_{2} (9 - 2^{x}) = 3 - x$
$t a k e a n t i l o g$
$9 - 2^{x} = 2^{3 - x} = 8. 2^{- x}$
$9 - 2^{x} = 8. 2^{- x}$
$m u l t i p l y 2^{x}, w e g e t a n d l e t 2^{x} = z$
$9 z - z^{2} = 8$
$z^{2} - 9 z + 8 = 8$
$(z - 1) (z - 8) = 0$
$z = 1, z = 8$
$2^{x} = 1, 2^{x} = 8$
$=> x = 0, x = 3$

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