Question

How do you solve `log_3x-2log_3 2=3log_3 3`?

Answer 1

I found : `x=108`

Explanation:

We can use the properties of logs that tell us:
`logx-logy=log(x/y)`
and also:
`logx^a=alogx`
in our case we get:
`log_3x-log_3(2^2)=log_3(3^3)`
then:
`log_3(x/2^2)=log_3(27)`
if the logs must be equal the arguments must be as well. So, we can write:
`x/4=27`
`x=4*27=108`