SOLUTION: What is x in the following equation? 125^(7x-2)=(1/625)^(9x-5)

Question 508900: What is x in the following equation?
125^(7x-2)=(1/625)^(9x-5)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i think i have an answer.
not sure if this is the way you're supposed to do it, but it appears to work.
at least i have an answer that appears to be only 1.5 * 10^-10 different which should well be within any rounding requirement you might have.

here's what i did:

125^(7x-2) = (1/625)^(9x-5)

125 = 5^3
(1/625) = 1/5^4 = 5^-4

the equation becomes:

(5^3)^(7x-2) = (5^-4)^(9x-5)

by the law of exponents that says that (x^m)^n = x^(m*n), the equation becomes:
5^(3*(7x-2)) = 5^(-4*(9x-5))
multiplying the exponents out gets us:
5^(21x-6) = 5^(-36x+20)
since the base is the same, this means that the equation is equal if and only if the exponents are equal.
this means that:
21x-6 = -36x+20
add 36x to both sides of this equation and add 6 to both sides of this equation to get:
21x + 36x = 20 + 6 which simplifies to:
57x = 26
divide both sides of this equation by 57 to get:
x = 26/57

that's your answer.

substitute for x in the original equation to get:
125^(7x-2) = (1/625)^(9x-5) becomes:
125^(7*26/57-2) = (1/625)^(9*26/57-5)
this simplifies to:
125^(68/57) = (1/625)^(-51/57)
this further simplifies to:
317.3780282 = 317.2780281
i stored the numbers internally and then subtracted them from each other.
this time the difference came out 0.
the reason is that i stayed with fractions until the end so there was no intermediate rounding error.
that appears to be your answer.
the solution was to get the bases to be the same and then apply the exponential law of (x^m)^n = x^(m*n).
since the equation was one of equality and the base was the same, this means that the exponents had to be equal which is how the problem was solved.

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