Question 804513
{{{1+log(8x)=16-2log(x)}}}
{{{1+log(8)+log(x)=16-log(x^2)}}}
{{{1.90+log(x)=16-log(x^2)}}}
{{{-14.1+log(x)=log(x^(-2))}}}
{{{-14.1=log(x^(-3))}}}
{{{-14.1=-3*log(x)}}}
{{{4.7=log(x)}}}
{{{10^(4.7)=x}}}
{{{x=50119}}}