Solve for x: 2^(3x - 2) = 4^x...I got it down to log4 = xlog2^(3x-2). Where to go from there?

Question

Thank you so much for your help in advance! I'm really having trouble with these summer assignments! :)

Answer 1

Last answer is correct but not the simplest method.

2^(3x - 2) = 4^x = (2^2)^x = 2^(2x)

3x - 2 = 2x

x = 2

Answer 2

2^(3x-2)=4^x

log base 2 to the 4^x=3x-2

x*log base 2 to the 4=3x-2

x*2=3x-2

2x=3x-2

-x=-2

x=2

Answer 3

2^(3x-2)=4^x=>

2^(3x-2)=2^(2x)=>

3x-2=2x=>

x=2

If you use "log",then

(3x-2)log2=xlog4=>

(3log2)x-(2log2)x=2log2=>

(log2)x=2log2=>

x=2

I think solve it in exponential equation is better