# Thread: is x to the power of 1/3 the same as x to the power of 2/6? see inside please

1. ## is x to the power of 1/3 the same as x to the power of 2/6? see inside please

x^(1/3) ... is it the same as x^(2/6) ?

lets consider x < 0

x^(1/3) is okay

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

then

x^(1/3) = (x^2)^(1/6) is okay since (-8)^2 = 64^(1/6) ((but this = 2))

and finally

x^(1/3) = (x^(1/6)^2) is not okay since (-8)^(1/6) = undefined

does anyone have an explanation, idea or comment about this? my math teacher brought this up to me today

2. Originally Posted by finch41
x^(1/3) ... is it the same as x^(2/6) ?

lets consider x < 0

x^(1/3) is okay

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

then

x^(1/3) = (x^2)^(1/6) is okay since (-8)^2 = 64^(1/6) ((but this = 2))

and finally

x^(1/3) = (x^(1/6)^2) is not okay since (-8)^(1/6) = undefined

does anyone have an explanation, idea or comment about this? my math teacher brought this up to me today
It depends if you are strict...I can see it from both ways...since you haev a brief indtroduction of i..but then it is elminated...so I would say they are the same..

3. Well, lets see, we have:

$x^{\frac{1}{3}}$

And we want to know if:

$x^{\frac{2}{6}}$

is the same.

Well, let's use the root versions of these:

$\sqrt[3]{x}$

and

$\sqrt[6]{x^2}$

Ok, let's use 8:

$\sqrt[3]{8} = 2$

$\sqrt[6]{64} = ?$

Well, we are going to assume that it is 2, well, what's 2 to the 6th power?

$2^6 = 2*2*2*2*2*2 = 4*2*2*2*2 = 8*2*2*2 = 16*2*2 = 32*2 = 64$

Looks like they're the same to me, now let's try -8

$\sqrt[3]{-8} = -2$

$\sqrt[6]{(-8)^2} = 2$

Looks like they're only the same for all $|x|$.

4. good posts guys

my teacher's point was that he was not sure if they are the same function since it can be seen that the domains are uncertain to be the same

if the domain of two functions is not the same then they are not the same function right

5. If you restrict the domain to $\mathbb{N}$ then they are the exact same function. When you cross into the negatives, then things get a bit funky.

$\{x^n = x^{\frac{2}{2}n}|\forall x\in\mathbb{N}\}$

6. could you explain what you mean by funky and by what you wrote there

7. What I mean by funky is that the two functions have opposite signs once your domain goes into the negative integers.

And the bottom just says that x to some power n is equal to x to the power of (2/2)n for all x in the natural number line (0,1,2,3,...).

As we saw, when $n=\frac{1}{3}$ and $x=8$ and $\frac{2}{2}n = \frac{2}{6}$, then:

$x^n = x^{\frac{2}{2}n}$ for all x > 0.