Question 856416
Logarithms of both sides should be equal, no matter the base of the logarithm taken.

{{{log(b,(x^2+3x-9)^(2x-8))=log(b,1)}}}
{{{(2x-8)*log(b,(x^2+3x-9))=0}}}
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The easier answer to find is to allow 2x-8=0, because either one factor OR the other must be zero.  Continuing with this factor, {{{highlight(x=-4)}}}------that's one answer.
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Using {{{log(b,(x^2+3x-9))=0}}}, and then translating into exponential form, you have 
{{{b^0=x^2+3x-9}}}
{{{1=x^2+3x-9}}}
{{{x^2+3x-10=0}}}-----Factorable!!!
{{{highlight_green((x-2)(x+5)=0)}}}
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ANSWER RESULTS:::::: {{{highlight(x=2)}}} or {{{highlight(x=-5)}}}, or as also found {{{highlight(x=-4)}}}.
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