Question 135305
Solve for x:
{{{2(8)^(4x)-33(8)^(2x)+16 = 0}}}
Let's temporarily substitute {{{z = (8)^(2x)}}} and {{{z^2 = (8)^(4x)}}}
{{{2z^2-33z+16 = 0}}} This can be solved for z by factoring.
{{{(2z-1)(z-16) = 0}}} and...
{{{2z-1 = 0}}} and {{{z-16 = 0}}}, so we have...
{{{2z = 1}}} so that:
{{{z = 1/2}}} and
{{{z = 16}}}
Now we replace the z with {{{z = (8)^(2x)}}}, so...
{{{(8)^(2x) = 1/2}}} but {{{8 = 2^3}}} and {{{1/2 = 2^(-1)}}}, so, making these substitutions, we get:
{{{(2^3)^(2x) = (2)^(-1)}}} Simplify the left side.
{{{(2)^(6x) = (2)^(-1)}}} Applying the rule: If{{{m^a = m^b}}} then {{{a = b}}}, so...
{{{6x = -1}}} Divide both sides by 6.
{{{x = -1/6}}} This is one of the roots.
{{{z = 16}}} Substitute {{{z = (8)^(2x)}}}
{{{(8)^(2x) = 16}}} Substitute: {{{8 = 2^3}}} and {{{16 = 2^4}}}
{{{(2^3)^(2x) = (2)^4}}} Simplify the left side.
{{{(2)^(6x) = (2)^4}}} or...
{{{6x = 4}}} Divide both sides by 6.
{{{x = 4/6}}} Simplify.
{{{x = 2/3}}} This is the second root.
Solution is: 
{{{x = -1/6}}}
{{{x = 2/3}}}

I'll leave the check for you to do.  It's not difficult and I've done it myself and the two solutions for x are indeed valid!