# How to process negative rational exponents

I have been doing questions for Precalculus fundamentals that include rational exponents, but they have all been positive up til this point, and I am a bit confused how to process them when they are negative. Do we just flip everything like with a normal negative exponent?

For example what I have been doing

$(4b)^{1/2}(8b)^{1/4} = 32b^{2/6}$

I have more examples if you wish I can post more.

But basically now I have a question that does this:

$(8y^3)^{-2/3}$ and I am supposed to simplify.

Another one $(u^4v^6)^{-1/3}$

I don't know how to make fractions and exponents to type these more properly/legibly.

Long story short: yes, with negative fractional exponents, everything is just "business as usual": all the rules you've learned apply equally well to all exponents, whether positive or negative, whole number or fraction.

For example, $b^mb^n = b^{n+m}$ whatever numbers $n$ and $m$ happen to be. (Be careful: this rule only applies if you're raising the same number to different powers; this rule doesn't apply to your first example, assuming I understand what you meant to write)

Similarly, $(b^n)^m = b^{nm}$, whatever kinds of numbers $n$ and $m$ are.

When we're talking about negative exponents, they are still defined as usual:

$$b^{-n} = \frac{1}{b^n}.$$ Similarly, fractional exponents are defined as usual: $$b^{m/n} = \sqrt[n]{b^m}.$$ For example, we can compute $b^{-2/3} = \frac{1}{b^{2/3}} = \frac{1}{\sqrt[3]{b^2}}$.

I hope this helps!

• Thank you your response helped! Commented Jan 27, 2014 at 14:57
• I'm very glad to hear it, I hope math.SE can help with any future math questions you encounter! If you're interested, take the tour to learn more about the website. Commented Jan 27, 2014 at 16:57
• That will be great, I knew there would be a better way to format my question but I was trying to review/answer it before class today. I will definitely take the tour and read how to use this site effectively before my next questions :D Commented Jan 27, 2014 at 23:53

I'm having trouble reading the questions, so I'll do a few examples:

$(x^3)^{-2}=x^{-6}=\frac{1}{x^6}$

$(\frac{3}{2})^{-\frac{2}{5}}=1\div(\frac{3}{2})^\frac{2}{5}=(\frac{2}{3})^\frac{2}{5}$

$2^{-\frac{1}{2}}=1\div 2^\frac{1}{2}=\frac{\sqrt{2}}{2}$

• I'm glad! Sorry I was unable to address your question directly - parsing is hard when it's not in LaTeX Commented Feb 5, 2014 at 4:57

Your first example should be $2^{7/4}b^{3/4}$, as $1/2 + 1/4 = 3/4$, $4^{1/2} = 2$, and $8^{1/4} = 2^{3/4}$. When taking exponents of exponents you distribute to the terms multiply the exponents:

$(x^a)^b = x^{ab}$

$x^{-a} = \frac{1}{x^a}$

$(8y^3)^{-2/3} = 8^{-2/3}y^{-2} = \frac{1}{4y^2}$
$(u^4v^6)^{-1/3} = u^{-4/3}v^{-2}$
• That's not true; if $b=1$ neither your nor the original version of the first equation is true. That's why I left the non-LaTeX there, because I had a feeling something has been lost in translation. Commented Jan 27, 2014 at 2:40
• Thank you for catching my mistake as I did make edit: several, but these are equations and hold true even if $b = 1$.