Are these exponential forms equal?

Question

Is $(\frac1{\sqrt x})^{11}$? the same thing as $x^{\sqrt{11}}$ ?

Basically what I'm asking is are those equivalent/the same?

Answer 1

$$\left(\frac{1}{\sqrt{x}}\right)^{11} = \left(x^{-1/2}\right)^{11} = x^{-11/2} \neq x^\sqrt{11} = x^{11^{1/2}}$$ So for equality you would need $\frac{-11}{2} = \sqrt{11}$ which is not possible.

Answer 2

Those are not the same. $\left(\frac1{\sqrt x}\right)^{11}=(x^{-1/2})^{11}=x^{-11/2}$.

Answer 3

We don't need an analytical approach to disapprove this equality. If we can find a counter example for $x\in\mathbb{R}^+$ where the left doesn't equal the right, we will be done. Take $x$ to be $4$. Then $\frac{1}{2^{11}}\ll 1$. The right hand side $4^{\sqrt{11}}\gg1$. So the equality doesn't hold. Another approach would be to take a significantly large $x$. Then the left hand side goes to zero and the right hand side goes to infinity.