Deeply confused about $\sqrt[5]{a^5}=(a^5)^{1/5}$

Question

So is this correct?

$\sqrt[5]{a^5} = \left(a^5\right)^{\frac{1}{5}}$

I need proof why $\left(a^5\right)^\frac{1}{5}$ can or cannot just be $a^\frac{5}{5}$ or just $a$?

I think of that rule of

$\left(a^m\right)^n = a^{m·n}$ and clearly $a = \left(a^5\right)^\frac{1}{5}$ is false

What rule, or what evidence can I always prove to myself to never have this confusion again?

so why can you not just multiply $\left(a^5\right)^\frac{1}{5}=a^\frac{5}{5}$? Cause fractions? Ok. I need confirmation.

If $\sqrt[5]{a^5} = \left(a^5\right)^{\frac{1}{5}}$ is correct then is $\left(a^x\right)^\frac{m}{n} = \sqrt[n]{a^x}^m$ likewise?

I hate confusion and making the same math mistakes twice.

truly appreciated

Answer 1

I think the reason you are confused is because you think you are wrong about something, when you are actually right!

$(a^5)^{1/5}$ is just $a$. In fact all of the expressions you list in your question are equal: $\sqrt[5]{a^5} = (a^5)^{1/5} = a^{5/5} = a^1 = a$.

You asked: "Why can you not just multiply..." and the answer is, you can just multiply.

HOWEVER: All of this comes with one huge warning. In general, this kind of manipulation only works if $a$ is positive. If $a$ is negative, and if the the exponents involved are even, weird things can happen.

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EDIT: Now that I've seen your most recent comment, I think the issue has nothing to do with exponents or roots... It has to do with parentheses. You asked:

My first thought is (32a^5)^1/5 = 5th SQRT(32a^5). Then I use the rule above, to get 32a^5/5 = 32a . Why is that usage wrong, and how can I clear future confusion, if the answer is 2a?

The problem is that you forgot that the fifth root operation applies to all of $32a^5$. So what you have is not $32a^{5/5} = 32a$, but rather $(32a^5)^{1/5}=2a$, because the fifth root of $32$ is $2$.

Answer 2

It is known that for $n \geq 0,m>0, a\geq 0$:

$$\sqrt[m]{a^n}=(a^{n})^{\frac{1}{m}}=a^{\frac{n}{m}}$$

So: $$\sqrt[5]{a^5}=a^{\frac{5}{5}}=a$$

Answer 3

Use the definition of $x^y$ for $x>0$, $x^y:=e^{y\ln x}$

For $a>0$,$$(a^5)^{\tfrac{1}{5}}=e^{\tfrac{1}{5}\ln(a^5)}=e^{\tfrac{5}{5}\ln(a)}=e^{\ln(a)}=a$$

For $a=0$, this definition can be continuously extended: $0^{\dfrac{1}{5}}=0$ (but not differentiable in $0$).

For $a<0$, a complex definition can be considered.

EDIT:

Not only is writing down the definition a good habit, it also makes it clear about where each variable live, which seems to be causing the problem.

Answer 4

Remember $$\sqrt[n]a: =a^{\frac1n},$$ $$(a^r)^s=a^{rs}.$$ where $n$ is an integer and $r, s$ are real. This is enough to handle all cases. (Actually, $n$ can be a real as well, but this is unusual with the root notation.)

So, $$\sqrt[n]{a^m}=(a^m)^{\frac1n}=a^{m\cdot\frac1a}=a^{\frac mn}=a^{\frac1n\cdot m}=(\sqrt[n]a)^m.$$

When $n=m$, all these equal $a^1=a$.

UPDATE: You also have $$(ab)^r=a^rb^r,$$ which you can apply with the root notation: $$\sqrt[n]{a^mb^m}=\sqrt[n]{(ab)^m}=(ab)^{\frac mn}=a^{\frac mn}b^{\frac mn}=\sqrt[n]{a^m}\sqrt[n]{b^m}=(\sqrt[n]a)^m(\sqrt[n]b)^m.$$