My Math Forum fraction in negative exponentiation fraction

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 September 6th, 2012, 08:41 AM #1 Newbie   Joined: Sep 2012 Posts: 11 Thanks: 0 fraction in negative exponentiation fraction I really don't know how to spell that correctly in english, but anyways. I know my fair share of fractions and exponentials, but I got stuck on this one: (1/-1/3 Where -1/3 is the exponential and 1/8 is just a normal fraction (is it called exponentials in english? for example: 2 in the exponential 5 is 32(because 2*2*2*2*2=32) So in words it should be something like (One over eight) in minus one third I hoped I managed to explain this properly, ps. I looked up the answer in the book and it turns out to be 2. But i still want to know how its done
 September 6th, 2012, 08:46 AM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: fraction in negative exponentiation fraction The word in English is simply exponent. Now, we are looking at $\left(\frac{1}{8}\right)^{-1/3}$. Using the laws of exponents, we know that the negative sign in the exponent means to take the reciprocal. So, $\left(\frac{1}{8}\right)^{-1/3}= 8^{1/3}$, since the reciprocal of $\frac{1}{8}$ is 8. Now, a fractional exponent means that you are taking a root of the number. Since the fraction is $\frac{1}{3}$, we are taking the third root (normally referred to in English as the cube root) of 8. Since $2^3= 8$, the cube root of 8 is 2, and that is your answer.
 November 2nd, 2012, 03:54 PM #3 Newbie   Joined: Nov 2012 From: Rio de Janeiro, Brazil Posts: 2 Thanks: 0 Re: fraction in negative exponentiation fraction $\left ( \frac{1}{8}\right )^{- \frac{1}{3}}=$ $\left ( \frac{8}{1}\right )^{\frac{1}{3}}=$ $8^{\frac{1}{3}}=$ $\sqrt[3]{8}=$ $\sqrt[3]{2^3}=$ $\fbox{2}$
November 4th, 2012, 06:05 AM   #4
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Re: fraction in negative exponentiation fraction

Hello, ThePope!

Quote:
 $\text{Simplify: }\:\left(\frac{1}{8}\right)^{-\frac{1}{3}}$

$\left(\frac{1}{8}\right)^{-\frac{1}{3}} \;=\;\left(\frac{1}{2^3}\right)^{-\frac{1}{3}} \;=\;\left(2^{-3}\right)^{-\frac{1}{3}} \;=\;2^{(-3)(-\frac{1}{3})} \;=\;2^1 \;=\;2$

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