Question

How do you solve for x in `log_3(x^(2)-2x)=1`?

Answer 1

`x_1=-1` and `x_2=3`

Explanation:

`log_3 (x^2-2x)=1`
`x^2-2x=3^1`
`x^2-2x-3=0`
`Delta = 4 +12 =16`
`x=(-b+-sqrt(Delta))/2=(2+-4)/2` => `x_1=-1` and `x_2=3`