Question 1186321
<br>
{{{2^(2x+2) - 33(2^x) + 8 = 0}}}<br>
Note the powers of 2 in the equation are 2x+2 and x.  When you need to solve an equation like this, you need to get the equation in the form of a quadratic by making the larger exponent twice the smaller.<br>
In this example, the smaller exponent is x, so you want to rewrite the equation with the larger exponent being 2x.  That is easy to do:<br>
{{{2^(2x+2)=(2^(2x))(2^2)=4(2^(2x))}}}<br>
So rewrite the equation as<br>
{{{4(2^(2x)) - 33(2^x) + 8 = 0}}}<br>
Then factor this as a quadratic with {{{2^x}}} as the variable:<br>
{{{(4(2^x)-1)(2^x-8)=0}}}
{{{4(2^x-1)=0}}} OR {{{2^x-8=0}}}<br>
(1) {{{4(2^x-1)=0}}}
{{{4(2^x)=1}}}
{{{2^x=1/4 = 1/2^2 = 2^(-2)}}}
{{{x = -2}}}<br>
(2) {{{2^x-8=0}}}
{{{2^x=8=2^3}}}
{{{x=3}}}<br>
ANSWERS: x=-2 or x=3<br>
CHECK:
(1) x=-2: {{{2^(2x+2)-33(2^x)+8=2^(-2)-33(2^(-2))+8=-32(2^(-2))+8=-32(1/4)+8=-8+8=0}}}<br>
(2) x=3: {{{2^(2x+2)-33(2^x)+8=2^8-33(8)+8=256-264+8=0}}}<br>