SOLUTION: I need not only answers,but detail solution too Find x when {{{5^(2x+2)}}}−{{{5^(x+6)}}} = {{{5^x }}}− 625

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Question 823881: I need not only answers,but detail solution too
Find x when 5%5E%282x%2B2%29−5%5E%28x%2B6%29 = 5%5Ex+− 625
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
5%5E%282x%2B2%29-5%5E%28x%2B6%29=5%5Ex+-+625
5%5E%282x%2B2%29-5%5E%28x%2B6%29=5%5Ex+-+5%5E4 (because 5%5E4=325 )
5%5E%28x%2B2%29%2A5%5Ex-5%5E%28x%2B2%29%2A5%5E4=5%5Ex+-+5%5E4 (because a%5Eb%2Aa%5Ec=a%5E%28b%2Bc%29 )
5%5E%28x%2B2%29%2A%285%5Ex-5%5E4%29=1%2A%285%5Ex+-+5%5E4%29 (taking out 5%5E%28x%2B2%29 as a common factor)
5%5E%28x%2B2%29%2A%285%5Ex-5%5E4%29-1%2A%285%5Ex+-+5%5E4%29=0 (subtracting the expression on the right hand side from both sides of the equal sign)
%285%5E%28x%2B2%29-1%29%2A%285%5Ex+-+5%5E4%29=0 (taking out %285%5Ex+-+5%5E4%29 as a common factor)
Since the product of two factors %285%5E%28x%2B2%29-1%29%2A%285%5Ex+-+5%5E4%29 is zero,
one of those two factors must be zero.
So either
5%5E%28x%2B2%29-1=0-->5%5E%28x%2B2%29=1-->x%2B2=0-->highlight%28x=-2%29
or
5%5Ex+-+5%5E4=0-->5%5Ex+=+5%5E4-->highlight%28x=4%29