How do you solve $log (5 (2^{x})) = log (3 - 2^{x + 2})$ $Log(5(2^x)) = log(3-2^(x+2))$ ?

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Answer 1 · 2016-04-22T06:44:10

$x = \frac{log 3}{log 2}$ $x = log 3/log 2$ /

Equating the arguments,

$5 (2^{x}) = 3 - 2^{x + 2} = 3 - 2^{x} 2^{2} = 3 - 4 (2^{x})$ $5( 2^x)=3-2^(x+2 )= 3-2^x2^2 =3-4( 2^x)$

So, $2^{x} = 3$ $2^x=3$ and.

$x log 2 = log 3$ $x log 2 = log 3$
$x = \frac{log 3}{log 2}$ $x=log 3/log 2$ .

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