How to read $( x^{2x^{-1/5}})^{-1}=\frac{1}{25}$

Question

I saw the following problem

$$\left( x^{2x^{-\frac{1}{5}}} \right)^{-1}=\frac{1}{25}$$

and I was curious whether I should read it as

$$\left( (x^{2x})^{(-\frac{1}{5})} \right)^{-1}$$

or $$\left( x^{(2x^{\frac{1}{5}})} \right)^{-1}$$

to clarify, am I to raise 2x to the -1/5 power first, or raise $x^{2x}$ to the -1/5 power?

$a^{b^c}$ means $a^{(b^c)}$ usually (for instance, when you are awake). — , Dec 28 '20 at 17:25
I would say it is $\left( x^{2\left(x^{-\frac{1}{5}}\right)} \right)^{-1}$ — ECL, Dec 28 '20 at 17:26
@ECL I agree with you. Simplifying it results in $x^x = 5^5$ — , Dec 28 '20 at 17:28
thanks that helps a lot, some of the confusion on my part might have been the typeface used in the original problem.... — Jesse, Dec 28 '20 at 17:41

score 0 · Answer 1 · edited Dec 28 '20 at 18:16

0

The first thing I would do is take the reciprocal of both sides: $x^{2x^{-1/5}}= 25$. From there, because you have "$x$" both as the base and in the exponent I expect you will need the "W function".

edited Dec 28 '20 at 18:16

answered Dec 28 '20 at 17:28

user247327

18,710

Please check your formatting again. – Dec 28 '20 at 18:01
Since $25=5^2,$ we can find the real valued solution without W. – Bumblebee Dec 30 '20 at 01:35

How to read $( x^{2x^{-1/5}})^{-1}=\frac{1}{25}$

1 Answers1