Is $\log(y^2)=e^{-x}$ a function of $x$?

Question

I need to determine whether $\log(y^2)=e^{-x}$ is a function of $x$, where $\log$ is the natural logarithm. And I have come up with two conflicting answers:

1. $\log(y^2)=e^{-x} \, \Rightarrow \, 2 \log(y)=e^{-x} \, \Rightarrow \, y=e^{\frac{e^{-x}}{2}}$
2. $\log(y^2)=e^{-x} \, \Rightarrow \, y^2=e^{e^{-x}} \, \Rightarrow \, y = \pm \sqrt{e^{e^{-x}}}$

From the (1) I would conclude it is a function and from (2) I would conclude that it is not. Where is my flaw?

Thanks!

Answer 1

They're the same thing. A fractional power is a root.

But your conclusion is sound. There are 2 values of y.