What is the solution to the equation below? log64x^2-log6x=2
x=1/2
x=3/2
x=3
x=9

First, we need to put the conditions: 4x^2 >0 which is obviously true
x>0
We will apply the formulas:
loga(x) - loga(y) = loga(x/y)
and
if loga(x)=b then x = a^b

log6(4x^2)-log6(x)=2
Applying first formula
log6(4x^2/x) = 2
log6(4x) =2

Applying the second formula:
4x = 36
x=9