How do I prove that 32^2x+1 = 8^3x-2

How do I prove that 32^2x+1 = 8^3x-2
hint: If x^a = x^b, then a=b
I have to find some way so that I can set (2x+1) equal to (3x-2) ...I think...Please help!!!

You don't really prove it, you just need to solve for the x value that makes the statement true.

First we need the bases to be equal, so ask yourself this: 8 to the what power gives 32.
8^y = 32
Take the natural log of both sides:
ln(8^y) = ln32
Rewrite
yln8 = ln32
y = ln32 / ln8
y = 5/3

Now we know that 8^(5/3) = 32 so we can change 32^(2x + 1) to 8^(5/3(2x + 1)) which means...

5/3(2x + 1) = 3x - 2
Solve for x
x = -11

To prove the equation, we need to express first the bases as powers of 2, hereby:
32=2^5 y 8=2^3, therefore:

2^5.(2x+1) = 2^3.(3x-2),
With which the exponents now are equal... Finally, we clear the value of "x".

5.(2x+1) = 3.(3x-2), or
10x+5 = 9x-6,
x = -11.

To verify it, we replace the value of "x" in the original proposition
32^2(-11)+1 = 8^3(-11)-2
32^-21 = 8^-35

2,465190329e-32 = 2,465190329e-32

32^(2x + 1) = 8^(3x - 2)

2^(10x + 5) = 2^(9x - 6)

10x + 5 = 9x - 6

x = -11

Try reducing the 32 to 2^5 and the 8 to 2^3 and then taking log\/2 of both sides