How do you solve ${log}_{6} (x + 19) + {log}_{6} (x) = 3$ $log_6(x + 19) + log_6(x) = 3$ ?

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Answer 1 · 2015-10-04T02:05:38

We must have $x > 0$ $x>0$ hence

$log_6(x + 19) + log_6(x) = 3=>log_6 x(x+19)=3=> x(x+19)=6^3=>x^2+19x-216=0=>x^2+27x-8x-216=0=> (x-8)*(x+27)=0=>x=8 or x=-27$

But $x=-27$ is rejected so $x=8$ is the only solution

How do you solve log_6(x + 19) + log_6(x) = 3log6(x+19)+log6(x)=3log_6(x + 19) + log_6(x) = 3?