# Inverse of a Fractional Exponent?

Hello Mathematics Stack Exchange,

I'm currently a Grade $$11$$ Math Student and to train for this year's exam I am going through a worked video of the previous years' exams to get a better understanding of what to expect.

I came across the following question:

$$\text{Consider the function}:$$ $$W=(x+2)^{\frac{2}{5}}$$ Make $$x$$ the subject of the formula. Seems pretty straightforward, right?

Well, this is the process they went through to isolate $$x$$:

$$W=(x+2)^{\frac{2}{5}}$$

$$x+2=W^{\frac{5}{2}}$$

$$x=W^{\frac{5}{2}}-2$$

So from what I can see, to get rid of the exponent $$\frac25$$ they raised the other side of the equation to the reciprocal which is $$\frac52$$. Now, I have searched online for 'inverse of a fractional exponent' etc, and I haven't really come across anything. Could someone explain to me more clearly why they raised it to the power of '$$\frac52$$'. I mean I am happy with accepting it as it is but I don't really know why they did this.

If there's something wrong with the question please let me know.

Cheers, Tom

• Recall that for positive reals $a$ you have that $(a^{b})^c = a^{(bc)}$, that $\frac{2}{5}\cdot \frac{5}{2}=1$ and that $a^1 = a$. For a more advanced note, recognize that I insisted on $a$ being a positive real. Be careful if the base is able to be negative or complex as the mentioned property may fail Sep 14, 2021 at 23:10

## 1 Answer

It comes from the exponent rule: $$(x^a)^b = x^{ab}$$. So if you want to "remove" an exponent, which means you're trying to get the final exponent to become $$1$$, you're trying to solve $$ab = 1$$ which means $$b = \frac{1}{a}$$. So if $$a = \frac{2}{5}$$, then $$b = \frac{5}{2}$$ meaning you have to raise everything to the power of $$\frac{5}{2}$$ to cancel out the power of $$\frac{2}{5}$$.

• My dad studied math in Uni and he's saying 'it's not possible' to do this... Weird, thanks for the answer; makes sense. Sep 14, 2021 at 23:16
• Note that if you have $y=x^\frac12$, then the inverse is not necessarily $y=x^2$, but rather $y=x^2,x\ge 0$. Also see roots of complex numbers which means that there are additional answers possibly. Sep 14, 2021 at 23:17