Question

How do you solve the equation `log_3(x+5)-log_3(x-7)=2`?

Answer 1

You can use the following facts:

`log_aM-log_aN=log(M/N)` and:

`log_ab=x => a^x=b`

So you get:

`log_3(x+5)-log_3(x-7)=2`
`log_3((x+5)/(x-7))=2`
`(x+5)/(x-7)=3^2`
`x+5=9(x-7)`
`x-9x=-63-5`
`-8x=-68`
`x=68/8=17/2`