SOLUTION: Solve for x: 32^x = 8^2x-1

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Question 978393: Solve for x: 32^x = 8^2x-1
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
32%5Ex%22%22=%22%228%5E%282x-1%29

No logarithms are needed, although you can get the answer
with logs.  But I won't use logs here:

Write 32 as 25 and 8 as 23

%282%5E5%29%5Ex%22%22=%22%22%282%5E3%29%5E%282x-1%29

Remove the parentheses by multiplying the exponents:

2%5E%285x%29%22%22=%22%222%5E%283%282x-1%29%29

The bases are equal, non-negative and not equal to 1, so
the exponents are equal, so we can drop the bases:

5x%22%22=%22%223%282x-1%29

You finish.

Edwin

Answer by ikleyn(52824) About Me  (Show Source):
You can put this solution on YOUR website!

As I understand,  you need to solve an equation

32%5Ex = 8%5E%282x-1%29,

correct?

If so,  then let us transform this equation step by step.
Since  32 = 2%5E5  and  8 = 2%5E3,  we have

32%5E%285x%29 = 2%5E%283%2A%282x-1%29%29.

It implies that

5x = 3*(2x - 1).

Solve this equation simplifying it step by step:

5x = 6x - 3,
-x = -3,
x = 3.

Check yourself the solution  x = 3.

Answer.  x = 3.