Question 481620
{{{2^(2x)-3*2^x-40=0}}}, substituting {{{2^x=y}}} we get the quadratic equation:

{{{y^2-3y-40=0}}}. We solve this equation by factoring:{{{(y-8)(y+5)=0}}}

The roots are: {-5, 8}. Now we find the roots of original equation:

{{{2^x=8}}}<=>{{{x=3}}} and {{{2^x=-5}}}<=>{{{x*ln2=ln(-5)}}}, since the log of 

negative numbers doesn't exist our equation has only one solution x=3.