Confused With Simplifying Exponents

Question

If you had $x^{2/4}$, would that simplify to $x^{1/2}$?

If you were to simplify $x^{2/4}$ to $x^{1/2}$, x cannot be negative for a real solution...

But if you left it as $x^{2/4}$, x would be able to be negative for a real solution, correct?

How do you know if $x^{2/4}$ = $(x^2)^{1/4}$ or $(x^{1/4})^2$?

Does $x^{2/4}$ = $x^{1/2}$ and would x have to be greater or equal to zero for a real solution?

Answer 1

But if you left it as $x^{2/4}$, $x$ would be able to be negative for a real solution, correct?

No. Most commonly rational exponents are defined as follows (this is from Michael Sullivan's College Algebra, 8th Edition):

Note that the definition assumes that the numerator and denominator in the exponent are "containing no common factors". This is highlighted in note (1): the fraction in the exponent "must be in lowest terms".

So: It is impermissible to transform $x^{2/4}$ into a radical expression without first reducing it to $x^{1/2}$. Allowing otherwise would generate the ambiguities that you're observing, i.e., it would fail to be well-defined.

Answer 2

So say we have some number $x \in \mathbb{R}$. Now we define a function that first takes the square root of our input, and then squares the result:

$$y(x) = (\sqrt{x})^2 = (x^\frac{1}{2})^2 = x$$

Notice that I can reduce the function down to just $x$ because the original domain is strictly any non-negative number due to the square root being first. Now switching up the order we have:

$$f(x) = \sqrt{x^2} = (x^2)^\frac{1}{2} \neq x$$

Here we cannot reduce it down to strictly $x$ because it does not hold true for $x < 0$. So the thing is, there is a fix to this situation. Observing the second function we are taking any real number and expecting back the same value with the condition that it is the positive version. In other words:

$$f(x) = |x|$$

This rule actually extends to any even root. Now for your question specifically, you have $g(x) = x^\frac{2}{4}$. So what we do is:

$$g(x) = x^\frac{2}{4} = (x^2)^\frac{1}{4} = ((x^2)^\frac{1}{2})^\frac{1}{2} = |x|^\frac{1}{2}$$

Of course doing it this way assumes that we are squaring first, but here the domain is all real numbers. On the other hand if you were to take the fourth root first, you would already have to restrict your domain to non-negative values right away. In which case, the absolute value would be unnecessary.