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I was just messing around with some exponential equations and stumbled upon this general expression.

If $x^n=y^z$, then $x=y^{z/n}$.

After further evaluation, I concluded that it is just a general method in which to find the root of two sides of an equation to find an unknown base (in this case $x$).

Blue
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iluvmath
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  • i am literally a high school student so i am still unaware of several maybe obvious concepts that may deem this result useless or incorrect for that matter. – iluvmath Sep 12 '22 at 19:35
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    This is not true. For example, $(-1)^2 = 1^{2}$, but $-1 \neq 1^{2/2}$. – TheSilverDoe Sep 12 '22 at 19:37
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 12 '22 at 19:37
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    looks fine :-) this is a nice (re)discovery! but the manupulations you have done do not always hold. can you find conditions under which your manupulations would work? – Math-fun Sep 12 '22 at 19:39
  • @TheSilverDoe what if one were find the absolute values of the bases like so: |-1|^2 = 1^2 and for your counter example |-1|=1^2/2 ? – iluvmath Sep 12 '22 at 19:40
  • @Math-fun your wording of rediscovery implies that something like this already exists yes? If possible could you give me the name of this property so i can read more into it. And yes lol i just found conditions in which the manipulation is indeed invalid – iluvmath Sep 12 '22 at 19:44
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    This is just a generic consequence of rules of exponents (which are taught in elementary algebra). – anon Sep 12 '22 at 19:47
  • let's say $y^z=a$ then $x^n=a$. By definition, $x$ is $n$-th root of $a$ – Vasili Sep 12 '22 at 19:54
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    It's true if you assume that all variables are positive real numbers. – Dan Sep 12 '22 at 19:55
  • Not quite a duplicate, but this may explain some things you are learning about exponents. https://math.stackexchange.com/questions/4412683/proofs-request-proofs-that-five-exponention-rules-hold-for-positive-real-bases/4412691#4412691 – Ethan Bolker Sep 12 '22 at 20:03
  • Welcome to MSE, and congrats on having fun finding something new to you. As others have pointed out, this is well-known, though it has limits: negative and complex numbers make it go all squirrelly and either untrue or indeterminate. – Eric Snyder Sep 12 '22 at 20:56

2 Answers2

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Your statement is

If $x^n=y^z$, then $x=y^{z/n}$.

Put some parentheses in the right places and we have

If $x^n=y^z$, then $\left(x^n\right)^{1/n}=x=\left(y^z\right)^{1/n}$

which follows from the laws of exponents $$\left(a^m\right)^n=a^{mn}.$$

This simply suggests that $(\cdot)^n$ and $(\cdot)^{1/n}$ are inverse operations when we have $(\cdot)^n$ well-defined.

Andrew Chin
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There are some conditions to consider for this equation...

  1. If $n=0$, we would have $x^0 = y^z \rightarrow 1 = y^z$, which implies $y = z = 1$. Conversely, If $z=0$, then $x^n = 1$, for which the only solution in real numbers would be $n=0$.

  2. If $y \lt 0$ and $n$ is even, the solution does not exist in the real numbers. E.g. if you have $x = -2^{1/2}$, it's equivalent to $x = \sqrt{-2}.$

To sum up, if $x^n = y^z$, then \begin{cases} x = y^{z/n}, & \text{$n \neq 0$} \\ x = 1, & \text{if $n=0$ and $y = z = 1$ OR if $n=0$ and $z=0$} \\ x = \text {undefined}, & \text{if $y<0$ and $n$ even} \end{cases}

bjcolby15
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