Question

2log[2](2)+2log[2](6)-log[2](3x)=3
x=2
x=6
x=16
x=18

Answer 1

2log[2](2)+2log[2](6)-log[2](3x)=3

There are a number of ways to solve this... first I see that log[2](2) = 1

2*1 +2log[2](6)-log[2](3x)=3

2log[2](6)-log[2](3x)=1

Next use nlog(a) = log(a^n)

log[2](36)-log[2](3x)=1

Then log(a) - log(b) = log(a/b)

log[2](36/3x) = 1
log[2](12/x) = 1

Now apply the definition of log base 2:

2^1 = 12/x

2x = 12
x=6

Should substitute into the original to verify that it works.

Answer 2

thanks