How do you solve $2^x^2 = 32(2^(4x))$ ?

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Answer 1 · 2015-10-26T21:27:25

$x= 5 or -1$

Looking at your format I assume you meant to write:

$2^ (x^2) = 32 (2^(4x))$

$=> (2^ (x^2))/(2^(4x)) = 32$

$=> 2^((x^2 - 4x)) = 32$

Taking logs:

$=>(x^2 - 4x)log(2) = log(32)$

$=> x^2 - 4x = (log(32))/(log(2))$

This gives us a quadratic of:

$x^2 - 4x - (log(32))/(log(2)) =0$

But $(log(32))/(log(2)) = 5$ giving:

$x^2 - 4x - 5=0$

$(x -5)(x+1)=0$

so $x= 5 or -1$
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Check

$x(x+1)-5(x+1)=x^2+x-5x-5$

$x^2-4x-5$

How do you solve 2^x^2 = 32(2^(4x))2^x^2 = 32(2^(4x))?