Question 1002287
<pre>
{{{2^(2x) - 2^(x+2) = 32}}}

Use the laws of exponents 
        {{{(a^b)^c=a^(bc)}}} and {{{a^b*a^c=a^(b+c)}}}
in reverse as
        {{{a^(bc)=(a^b)^c}}} and {{{a^(b+c)=a^b*a^c}}}

to rewrite the terms on the left of

{{{2^(2x) - 2^(x+2) = 32}}}

as

{{{(2^x)^2 - 2^x*2^2 = 32}}}

{{{(2^x)^2 - 2^x*4 = 32}}}

{{{(2^x)^2 - 4(2^x) = 32}}}

{{{(2^x)^2 - 4(2^x)- 32=0}}}

Let u = 2<sup>x</sup>

{{{u^2-4u-32=0}}}

Factor the left side:

{{{(u-8)(u+4)=0}}}

u-8 = 0;   u+4 = 0
  u = 8;     u = -4

Since u = 2<sup>x</sup>

2<sup>x</sup> = 8;  2<sup>x</sup> = -4

The second has no real solution so we ignore it.
 
2<sup>x</sup> = 8

2<sup>x</sup> = 2<sup>3</sup>

The bases are equal and positive and not 1, so 
the exponents are equal.

x = 3

Edwin</pre>