Question

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

Answer 1

`x=1/9`

Explanation:

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`