h3ndu
Registered User regular

Ok - so, I have a final tomorrow in math 108, intermediate algebra - I have an A in the class, and I just took the practice final online - I did well, but there was one problem that stuck me - I searched through all my notes and I couldn't find anything on it - help me please?

the problem - (125/8) [so, one hundred and twenty five over eight] to the -2/8 th power. The answer to the problem is 4/25 [four over twenty five].

If anyone could walk me through this I would be very appreciative.

the problem - (125/8) [so, one hundred and twenty five over eight] to the -2/8 th power. The answer to the problem is 4/25 [four over twenty five].

If anyone could walk me through this I would be very appreciative.

Lo Que Sea, Cuando Sea, Donde Sea.

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## Posts

If so, X^(2/3) is best computed as X^(1/3)^2, so take the cuberoot and square it. So 125/8 to the -2/3 would be 8^(2/3) / 125^(2/3) = 2^2/5^2 = 4/25.

PowerpuppiesonAnyway, for how to do that kind of thing:

1) What do you do with negative exponents?

2) What do you do with fractions raised to a power?

enlightenedbumonno - no, it was definetly (125/8) to the -2/8.

I just ... seeing it in line form kind of sucks.

''''''''''-2/8

125

8

that's the problem - and when I finished the web test it told me the answer was

4

---

25

h3nduonI mean I dunno what to tell you man. The test is wrong. 125/8^(-2/3) is 4/25. 125/8^(-2/8) is not, plus what kind of math problem starts with easily reducible fractional exponents? Some sort of typo or something.

PowerpuppiesonI must have misread it and then tried to solve. I can't imagine the test itself would be wrong. It has to be an error on my part. It is kind of late right now.

Thanks for your help.

h3nduonSkyCaptainon