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 ThePope September 6th, 2012 08:41 AM

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:

(1/8)-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

 icemanfan September 6th, 2012 08:46 AM

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.

 danjr5 November 2nd, 2012 03:54 PM

Re: fraction in negative exponentiation fraction

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

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

$8^{\frac{1}{3}}=$

$\sqrt[3]{8}=$

$\sqrt[3]{2^3}=$

$\fbox{2}$

 soroban November 4th, 2012 06:05 AM

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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