Solve $(3^x - 1)(2^(2x) - 1/16) = 0$ for x?

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Answer 1 · 2017-08-27T14:26:12

$x=-2$
$x=0$

$color(blue)("Determine the first root")$

Before we start: $log(a^b)$ is the same as $blog(a)$

Divide both sides by $(3^x -1)$

Note that $0/(3^x-1)=0$

$2^(2x)-1/16=0$

$2^(2x)=1/16$

Take logs of both sides. I choose $log_10$

$2xlog(2)=log(1/16)$

$x=(log(1/16))/(2log(2))$

$x=-2$
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Check
$(3^x - 1)(2^(2x) - 1/16) = 0" " ->" " (3^(-2)-1)(2^(2xx(-2))-1/16)=0$

$color(white)("bbbbbbbbbbbbbbbbbbbbbs") ->" "-8/9(1/16-1/16)=0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Determine the second root")$

Set $(3^x-1) =0$

$3^x=1$

but as $3^0=1$ we must have

$x=0$

Tony B

Solve (3^x - 1)(2^(2x) - 1/16) = 0(3^x - 1)(2^(2x) - 1/16) = 0 for x?