Question 219792
{{{logx+log((x-11))=log((8x))}}}
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logs are added when their arguments are multiplied.
log(x) + log(x-11) = log(x*(x-11))
{{{log((x^2 - 11x)) = log(8x)}}}
{{{x^2 - 11x = 8x}}} since the logs are equal, the arguments are equal
{{{x^2 - 19x = 0}}}
x = 0  log(0) isn't valid, so ignore this solution
x = 19