Question

How do you solve `log_3 (2x-4) - log_3 (x-1)=log_3 (x-2)`?

Answer 1

I found:
`x=3`

Explanation:

We can use the property of logs:
`logx-logy=log(x/y)`
And write:
`log_3[(2x-4)/(x-1)]=log_3(x-2)`
If the logs are equal the arguments must be as well, so:
`(2x-4)/(x-1)=x-2`
`2x-4=(x-1)(x-2)`
`2x-4=x^2-2x-x+2`
`x^2-5x+6=0`
Use the Quadratic Formula:
`x_(1,2)=(5+-sqrt(25-24))/2=`
Two solutions:
`x_1=3` YES
`×_2=2` NO