Question 983580

{{{log((x + 8)) = 1 + log((x - 10))}}}... since base is {{{10}}}, we can write {{{1}}} as {{{log((10))}}}

{{{log((x + 8)) = log((10)) + log((x - 10))}}}

{{{log((x + 8)) = log((10(x - 10)))}}}...if log same, we have

{{{(x + 8) = 10(x - 10)}}}...solve for {{{x}}}

{{{x + 8 = 10x - 100}}}

{{{100+8 = 10x-x}}}

{{{108 = 9x}}}

{{{x=12}}}