Help with logs?

Question

Ok so im getting the problem down to... 3xlog4 = (x-.5)log16 also xlog9 = (3x-2)log 3

you just have to answer one , i just need to see how its done >.< so explanations are nice...

Answer 1

log16 = log 4^2 = 2 log 4
log 9 = log 3^2 = 2 log 3

so
3 log 4 = 2(x - 5)log4
3 = 2x - 10
x = 13/2

if 3 x log 4 = 2 (x - 5) log4
3x = 2x - 10
x = -10

x log 9 = (3x -2) log 3 = 2 x log 3
3x -2 = 2x
x = 2

Answer 2

Another way to solve it:

log9 = log(3^2) = 2log3

therefore
2xlog3 = (3x-2)log3

The log3 on both sides cancel leaving

2x = 3x-2
2 = 3x-2x = x

Thus x =2

Answer 3

Let's start with 3xlog4 = (x-.5)log16.

3xlog4 = (x-.5)log16
3xlog4 = (x-.5)log(4^2)

But log(4^2) = 2log4, so

3xlog4 = 2(x-.5)log4 = (2x - 1)log4 = 2xlog4 - log4

Now just use algebra to solve for x.

xlog4 = -log4
x = -1