How do you solve $log_8x+log_8(x+6)=log_8(5x+12)$ ?

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Answer 1 · 2016-07-12T07:50:47

$:. x=3$

$log_8x + log_8(x+6)= log_8(5x+12)$

$=>log_8(x*(x+6))= log_8(5x+12)$

$=>x(x+6)= 5x+12$

$=>x^2+6x-5x-12=0$

$=>x^2+4x-3x-12=0$

$=>x(x+4)-3(x+4)=0$

$=>(x+4)(x-3)=0$

$:.x=-4 and x=3$

But x=-4 is not possible bacause
$log("-ve number")$ is undefined.

How do you solve log_8x+log_8(x+6)=log_8(5x+12)log_8x+log_8(x+6)=log_8(5x+12)?