Question 712630
Common logarithms not necessary, pointless.   See how this solution works:


{{{5^x+125(5^(-x))=30}}}, as given.
{{{5^x+125/(5^x)-30=0}}}
Multiply both sides by 5^x.
{{{(5^x)5^x+125*(5^x)(5^(-x))-30*5^x=0*5^x}}}
{{{5^(2x)-30*5^x+125=0}}}, which is factorable.
{{{highlight((5^x-5)(5^x-25)=0)}}}


Taking {{{5^x-5=0}}}, 5^x=5, {{{x=1}}}.


Taking {{{5^x-25=0}}}, 5^x=25, 5^x=5^2, {{{x=2}}}.


Answer: {{{highlight(x=1)}}} or {{{highlight(x=2)}}}.