# If 2^(x + 8 - x^2)/(4^(3x - 2) = 16, what is x.

Posted on

The equation `2^(x + 8 - x^2)/(4^(3x - 2)) = 16` has to be solved for x.

`2^(x + 8 - x^2)/(4^(3x - 2)) = 16`

=> `2^(x + 8 - x^2)/(2^(6x - 4)) = 16`

=> `2^(x + 8 - x^2 - 6x + 4) = 16`

=> `2^(x^2 - 5x + 12) = 2^4`

As the base is the same equate the exponent.

x^2 - 5x + 12 = 4

=> x^2 - 5x + 8 = 0

This quadratic equation only has complex roots, `x = (5 +-sqrt(25 - 32))/2 = 5/2 +- (i*sqrt 7)/2`

Posted on

Sorry about the typo in the response.

`2^(x + 8 - x^2 - 6x + 4) = 16`

=> `2^(x + 8 - x^2 - 6x + 4) = 2^4`

Equate the exponent

`x + 8 - x^2 - 6x + 4 = 4`

=> `-x^2 - 5x + 8 = 0`

=> `x^2 + 5x - 8 = 0`

=> `x = (-5 +- sqrt(25 +32))/2`

=> `x = -5/2 +- sqrt(57)/2`

Posted on

Given

`2^(x+8-x^2)/4^(3x-2)=16`

`2^(x+8-x^2)/(2^2)^(3x-2)=2^4`

`2^(x+8-x^2)/2^(6x-4)=2^4`

`2^((x+8-x^2)-(6x-4))=2^4`

`2^(-x^2-5x+12)=2^4`

`=>-x^2-5x+12=4`

`=>-x^2-5x+8=0`

`=>x^2+5x-8=0`

`x=(-5+-sqrt(25+32))/2`

`x=(-5+-sqrt(57))/2`

So x has real values and equal to `(-5+-sqrt(57))/2`