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 $x$:




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

  • 1
    $\begingroup$ 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 $\endgroup$
    – JMoravitz
    Commented Sep 14, 2021 at 23:10

1 Answer 1


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 $\frac{2}{5}$.

  • $\begingroup$ My dad studied math in Uni and he's saying 'it's not possible' to do this... Weird, thanks for the answer; makes sense. $\endgroup$ Commented Sep 14, 2021 at 23:16
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Sep 14, 2021 at 23:17

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