Question

How do you solve `2log_3(x+4)=log_3(9)+2`?

Answer 1

I found `x=5`

Explanation:

we can use a little trick and write `2=log_3(9)`
so you get:
`2log_3(x+4)=log_3(9)+log_3(9)`
`cancel(2)log_3(x+4)=cancel(2)log_3(9)`
for the logs to be equal the arguments must be equal so:
`x+4=9`
`x=5`