log[3](2x+8)+log[3](3x)=log[2](3)2

Question

Looking to solve equation with both the exact and decimal approximation

Answer 1

log[3](2x + 8) + log[3](3x) = log[2] (3)^2

rule:  use log[b] x^n = n log[b] x

log[3](2x + 8) + log[3](3x) = 2 log[2] 3

rule:  use log [b] a = 1/(log [a] b)

log[3](2x + 8) + log[3](3x) = 1/(2 log[3] 2)

I expanded and got as far as I can... hope this helps