Question 972085
{{{8 = 4^(x^2)*2^(5x)}}}


{{{2^3=(2^2)^(x^2)*2^(5x)}}}


{{{2^3=(2^(2x^2))*2^(5x)}}}


{{{2^3=2^(2x^2+5x)}}}



Same base of 2 on both sides, so their exponents are equal:


{{{3=2x^2+5x}}}------More easily solvable in this form now.




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You should at least obtain a quadratic equation equivalent to the one shown there.


{{{2x^2+5x-3=0}}}
discriminant is {{{25+8*3=25+24=49=7^2}}}.
The quadratic part is factorable.(?)
{{{x=(-5+- 7)/4}}}
x=-3  or  x=1/2


or,
{{{(2x-1)(x+3)=0}}}