Question 331640
{{{ 2^(2x) - 3 * 2^x - 40= 0 }}}

Let {{{y=2^x}}}  then {{{Y^2=(2^x)^2 = 2^(2x)}}}

substitute in problem equation

{{{Y^2 - 3*y -40 =0}}}

factor this equation

(Y-8)*(y+5)=0

either y=8 or y=-5
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Case y=8

y = {{{8 = 2^x}}} solve for x   
{{{ln8 = ln(2^x) = x*ln2}}}
ln8/ln2 = x
3=x
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Case y=-5
y = {{{-5 = 2^x}}}
ln(-5)=x*Ln2   STOP!!!!  ln(-5) does not exit, log exist only for values greater than 0
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so there is only one answer: X=3