I had a question in my calculus book that was written a little strangely and I tried to rewrite it to be more simple but it did not work out. It made me realize something about powers and radicals though.

If I have $\sqrt{2^2}$ that is like $2^{2* \frac{1}{2}}$ if I am correct. I was thinking that I could cancel them out somehow but I doesn't really seem to be useful ever because they are reverse operations. For example if I have something more complex like

$2^{3*\frac{1}{5}}$ I can't really cancel anything out can I? Even though it is like I am multiplying $2*2*2$ and also diving by $2*2*2*2*2$ I can only make it a power of $\frac{3}{5}$ correct?

I guess what I am asking is that there is no simpler way to do work with this is there? If I have a power and a radical they aren't applied similar to $2^2 * 2^5$ where I just add the powers, I actually have to multiply powers because it is more like (for the power and radical example) $(2^2)^ \frac{1}{5}$ where I would then multiply out the powers.

  • 4
    $\begingroup$ There is nothing wrong with "cancelling them out" and writing $\sqrt{2^2} = 2$. If instead you have $\sqrt{x^2}$ you can write it as $|x|$. Also, what you said about $2^{3/5}$ is incorrect. It is not like "multiplying by $2*2*2$ and dividing by $2*2*2*2*2$." That would give you $2^{3}2^{-5} = 2^{3-5} = 2^{-2}$. There's nothing you can do to simplify $2^{3/5}$, except maybe write it as $(2^3)^{1/5}$ or $(2^{1/5})^3$. $\endgroup$ Feb 19 '12 at 21:13
  • $\begingroup$ If you are more interested in the knitty griddy complex analysis side of this issue, you should look it up. (P.S., try to find ALL solutions to $x^2=1$ by subtracting and using quadratic formulae, that should get you thinking) $\endgroup$ Jan 9 '16 at 15:26

The squaring function, $x\mapsto x^2$, and the square root function, $x\mapsto\sqrt{x}$, are not exactly inverses, because $\sqrt{x^2}$ does not always equal $x$; for instance, $\sqrt{(-1)^2} = \sqrt{1} = 1 \neq -1$.

The square root is the inverse of the "restricted squaring function", the function that is defined only for nonnegative $x$ and is defined by the rule $f(x) = x^2$, $x\geq 0$.

So in you instance, where your $x$ is positive, they are inverses of each other. But for an arbitrary, unknown quantity $a$, you have to be careful.

While $\left(\sqrt{a}\right)^2 = a$ is always true (because for the left hand side to even make sense you need $a\geq 0$), $\sqrt{a^2} = a$ is only true if $a\geq 0$; if $a\lt 0$, then both sides make sense but they are different. That's why the "correct" way to cancel the square root with the regular (unrestricted) squaring function is via the equalities $$\left(\sqrt{a}\right)^2 = a,\qquad \sqrt{a^2} = |a|.$$

You generally run into similar problems with fractional exponents. For example, $(-1)^1 = -1$; but what is $(-1)^{2/2}$? It "should" be the same as $((-1)^2)^{1/2}$, which is $1$. So you do have to be careful with radicals. So long as you stick with nonnegative bases, you can use the regular rules of exponents even with rational (or irrational!) exponents. If you are not positive that your base is nonnegative, then you should be very, very careful.


We know the rules of exponents (up to some nitpicky stuff to keep zeros out of the denominator and avoiding imaginary numbers and whatnot) are:

$a^b a^c = a^{b+c}$
$(a^b)^c = a^{bc}$

For "most" real $a, b, c$.

Two concrete examples pertaining to your question:

$\sqrt{x} * x^2 = x^{1/2}*x^2=x^{2 + 1/2} = x^{5/2}$

$\sqrt[4]{x^{16}} = (x^{16})^{1/4}=x^{16/4}=x^{4}$

Notice, in the first one, they were "applied" by adding the powers, and we used the first property I listed. In the second, the powers are multiplied because we're using the second property. If you would like more explanations or examples, I'd be glad to supply.

In summation: The simplest way to do the work is to be careful and know which rules get applied when. This mainly means solving some problems to get you familiar with the notation. For completeness, here's a link to the exponent laws.

  • $\begingroup$ See Arturo's answer for the gorier details I avoided! :) $\endgroup$
    – Tyler
    Mar 21 '12 at 4:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy