Is $8^x$ positive or negative?

Question

What is wrong with this logic?

$8^x = (\sqrt{8}^2)^x = (\sqrt{8}^x)^2 \Rightarrow 8^x \geq 0 \forall x$

Edit: Clarification, the answer to $8^{1/2}$ permits both $\sqrt{8}$ and $-\sqrt{8}$ (hence my confusion). That would also suggest $8^x$ is not a well defined function?

Answer 1

What is confusing you is that $9^{1/2}=3$ and $9^{1/2}\ne -3$.

We want the square root ($x^{1/2}$) to be a function, for this purpose you should get one and only one answer for any $x$. This means that "The square root of $9$ is a number whose square is $9$" is wrong, because then you can get different answers depending on who you ask, you always want to get the same answer. Which means you need to pick one of the numbers to be the square root of $9$. This is a convention that we prefer the nonnegative solution to the nonpositive.

Answer 2

Assuming $x$ is a real number, nothing is wrong with it. The conclusion is true.

Answer 3

What is special about the number 8 ? Do you know that $a^x=e^{x\log a}>0$ for all $a>0$?

Simply the exponential function is positive.