How do you solve $log_b9+log_bx^2=log_bx$ ?

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Answer 1 · 2016-07-25T09:53:34

$x=1/9$

Since $loga+logb=log(ab)$ , you can write

$log_b 9x^2=log_b x$

that's equivalent to:

$9x^2=x and x>0$

and the solution is:

$x=1/9$

How do you solve log_b9+log_bx^2=log_bxlog_b9+log_bx^2=log_bx?