Question

How do you solve `log_3 x - log_3 4 = log_3 12`?

Answer 1

It is

# log_3 x - log_3 4 = log_3 12=>log_3(x/4)=log_3(12)=> x/4=12=>x=48#

Answer 2

I found: `x=48`

Explanation:

We can start by using the property of the logs on subtractions to get a ratio of the arguments and write:
`log_3(x/4)=log_3(12)`
raise both side as exponents of `3`:
`3^(log_3(x/4))=3^(log_3(12))` this allows us to cancel out `3` and `log_3` to get:
`cancel(3)^(cancel(log_3)(x/4))=cancel(3)^(cancel(log_3)(12))`
`x/4=12`
`x=48`