Solve: $2=log_6x + log_6 (x+9)$ ?

Question

1419 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-22T02:42:30

$x=3$

$2=log_6x + log_6 (x+9)$

Remember $log_6 6= 1$

$:.2log_6 6 = log_6 x + log_6 (x+9)$

Also remember that $log_a b + log_a c = log_a (bxxc)$

$:. 2log_6 6 = log_6 (x(x+9))$

$log_6 6^2 = log_6(x^2+9x)$

$:. 36= x^2+9x$

$x^2+9x-36 =0$

$(x+12)(x-3)=0 -> x=-12 or x=3$

But $log_6 x$ is only defined for $x>0$

$:. x=3$

Solve: 2=log_6x + log_6 (x+9)2=log_6x + log_6 (x+9) ?