Understanding Rational Exponents with Negative Bases

AI Thread Summary
The discussion focuses on the complexities of evaluating expressions of the form (a^n)^(1/m) when a is negative. Participants highlight that such expressions can yield different results depending on the order of operations, specifically contrasting (a^n)^(1/m) with a^(n/m) and (a^(1/m))^n. A key point raised is that the absolute value may be necessary in simplifications, particularly when n and m are positive even integers. The conversation emphasizes the lack of clear rules for handling negative bases in these scenarios, leading to confusion among learners. Overall, the topic underscores the need for more resources to clarify these mathematical principles.
bloodasp
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Can anyone point me to a text or link that summarizes the rules when evaluating/simplifying an expression of the form
(a^n)^(1/m) for a < 0. (a^n)^(1/m) yields different answers for a^(n/m) and (a^(1/m))^n.

Ex:

(-8)^(2/6) = (-8)^(1/3) = -2
(-8)^(2/6) = ((-8)^2)^(1/6) = 2
(-8)^(2/6) = (-8^(1/6))^2 = undefined

Thank you very much!
 
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The general function, ax, is only defined for positive a.
There simply are no ways of giving "rules" for such manipulation with a negative.
 
Thanks HallsofIvy

i've read in some book that
(a^n)^{1/m}
where a < 0, n and m are positive even integers and a^{1/m} is defined can be simplified to
|a|^{n/m}

There are just a few examples on this subject that's why I'm looking for other resources. I've solved several exercises and I different answers. I miss out on when to place the absolute value bars and when not to. As I understand it, the absolute value bars may be removed if n/m always yields a positive value for |a|^{n/m}, otherwise, the absolute value bars must be retained.

I know this is elementary for you guys. :biggrin:
 
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Just a LaTex note. surround your exponent with curly brackets {} to get it all elevated.\

a ^{ \frac 1 m}

click on the equation to see the code.
 
Just a rewrite

bloodasp said:
Can anyone point me to a text or link that summarizes the rules when evaluating/simplifying an expression of the form
(a^n)^(1/m) for a < 0. (a^n)^(1/m) yields different answers for a^(n/m) and (a^(1/m))^n.

Ex:

(-8)^(2/6) = (-8)^(1/3) = -2
(-8)^(2/6) = ((-8)^2)^(1/6) = 2
(-8)^(2/6) = (-8^(1/6))^2 = undefined

Thank you very much!

Can anyone point me to a text or link that summarizes the rules when evaluating/simplifying an expression of the form
(a^n)^{1/m} for a < 0. (a^n)^{1/m} yields different answers for a^{n/m} and (a^{1/m})^n.

Ex:

(-8)^{2/6} = (-8)^{1/3} = -2
(-8)^{2/6} = ((-8)^2)^{1/6} = 2
(-8)^{2/6} = ((-8)^{1/6})^2 = undefined

Thank you very much!
 
Last edited:
bloodasp said:
i've read in some book that
(a^n)^{1/m}
where a < 0, n and m are positive even integers and a^{1/m} is defined can be simplified to
|a|^{n/m}
No, that's not true. There's no such way to simplify that. :frown: :smile:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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