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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.

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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!

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  • $\begingroup$ Thank you your response helped! $\endgroup$
    – Jennifer
    Jan 27 '14 at 14:57
  • $\begingroup$ 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. $\endgroup$
    – pjs36
    Jan 27 '14 at 16:57
  • $\begingroup$ 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 $\endgroup$
    – Jennifer
    Jan 27 '14 at 23:53
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I'm having trouble reading the questions, so I'll do a few examples:

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

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

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

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  • $\begingroup$ I'm glad! Sorry I was unable to address your question directly - parsing is hard when it's not in LaTeX $\endgroup$ Feb 5 '14 at 4:57
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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} $

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

So for your problems

$(8y^3)^{-2/3} = 8^{-2/3}y^{-2} = \frac{1}{4y^2} $

$(u^4v^6)^{-1/3} = u^{-4/3}v^{-2} $

Use MathJaX to make the expressions more legible. This website has some good guides.

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  • $\begingroup$ 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. $\endgroup$
    – pjs36
    Jan 27 '14 at 2:40
  • $\begingroup$ Thank you for catching my mistake as I did make edit: several, but these are equations and hold true even if $b = 1$. $\endgroup$
    – qwr
    Jan 27 '14 at 2:48
  • $\begingroup$ Thank you your response helped! $\endgroup$
    – Jennifer
    Jan 27 '14 at 14:55

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