SOLUTION: Solve for x if x:3^2x = 9^2x-1

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Question 1177146: Solve for x if x:3^2x = 9^2x-1
Found 2 solutions by greenestamps, MathLover1:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


3%5E%282x%29=9%5E%282x%29-1

Rewrite 9 as 3^2 to make the bases the same:

3%5E%282x%29+=+%283%5E2%29%5E%282x%29-1
3%5E%282x%29+=+3%5E%284x%29-1
3%5E%284x%29-3%5E%282x%29-1+=+0

This is a quadratic equation with "3^(2x)" as the "variable". To make it easier to solve, let y=3^(2x), so that 3^(4x) = y^2:

y%5E2-y-1=0

That quadratic does not factor; using the quadratic formula gives us

y+=+%281%2B-sqrt%285%29%29%2F2

Those values are one positive and one negative. We can ignore the negative root, because y=3^(2x) is never negative. So

y+=+3%5E%282x%29+=+%281%2Bsqrt%285%29%29%2F2
2x%2Alog%283%29+=+log%28%28%281%2Bsqrt%285%29%29%2F2%29%29
x+=+%28%28log%28%281%2Bsqrt%285%29%29%2F2%29%29%29%2F%282%2Alog%283%29%29

A calculator shows the value to be 0.219 to 3 decimal places.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

3%5E%282x%29+=+9%5E%282x-1%29
3%5E%282x%29+=+%283%5E2%29%5E%282x-1%29
3%5E%282x%29+=+3%5E%282%282x-1%29%29
3%5E%282x%29+=+3%5E%284x-2%29............if base same, exponents are same too
2x+=+4x-2
2+=+4x-2x
2+=+2x
x+=+1