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Question
Solve for x:
`(9/25)^(1 - 2x) = (125/27)^(x - 2)`
Sum
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Solution
We need to solve:
`(9/25)^(1 - 2x) = (125/27)^(x - 2)`
Step 1: Prime factorization
- `9 = 3^2, 25 = 5^2 ⇒ 9/25 = (3/5)^2`
- `125 = 5^3, 27 = 3^3 ⇒ 125/27 = (5/3)^3`
So, `(9/25)^(1 - 2x) = (3/5)^(2(1 - 2x)`
`(125/27)^(x - 2) = (5/3)^(3(x - 2)`
Step 2: Rewrite bases
`(3/5)^(2(1 - 2x)) = (5/3)^(-2(1 - 2x))`
So equation becomes:
`(5/3)^(-2(1 - 2x)) = (5/3)^(3(x - 2))`
Step 3: Equating exponents
Since bases are equal:
`-2(1 - 2x) = 3(x - 2)`
Expand: `-2 + 4x = 3x - 6`
`4x - 3x = -6 + 2`
`x = -4`
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