If $2^x=0$, find $x$. 
If $2^x=0$, find $x$.

Solution: I know range of $2^x$ function is $(0,\infty)$. 
So $2^x=0$ is not possible for any real value of $x$
Hence, equation is wrong. We can't find value of $x$. Am I right?
Please help me.
Can $x$ be in $[-\infty,\infty]$? 
i.e is $2^x=0$ possible for $x=-\infty$?
 A: There is no solution for this equation, note that's why log x is defined only to $x\gt 0$.
See the graph of $2^x$ and see that this function is never $0$:

Remark
The graph is not a proof this function is never $0$, it's just to illustrate what others have said in another answers graphically.
A: Yes. You are right, there is no $x \in \mathbb{R}$ such that $2^x = 0$. However, note that
$$\lim_{x \to -\infty} 2^x = 0$$
However, on the extended real line, i.e., $\mathbb{R} \cup \{-\infty\} \cup \{\infty\}$, there exists a solution, since one of the ways of defining "$2^{-\infty}$" on the extended real line is to define "$2^{-\infty}$" as the limit of $\lim_{x \to -\infty} 2^x$, which is $0$.
A: Let $x$ be a real number $x>0$, and consider $w=2^x$ and $z=2^{-x}$. Then $wz=1$. But then neither of them can be $0$, since in either case we would reach the absurdity that $0=1$. Thus, by symmetry, and since $2^0=1$, we see that $2^x\neq 0$ for each $x\in \Bbb R$, that is, $2^x=0$ is unsolvable in $\Bbb R$. The very same proof applies when $z\in\Bbb C$.
The fact that "the range of the function $2^x$ is $(0,+\infty)$" is true because of the above, and not conversely. It is true we may define $2^x=0$ when $x=-\infty$ to extend $2^x$ continuously to $\Bbb R^*=\Bbb R\cup\{-\infty,+\infty\}$, but there is not much more to it than that.
A: Originally people did not have the concept of "negative numbers." Similarly there were no solutions for $x^2 = -1$ which nowadays we have $+i,-i$ as solutions. If you can come up with a number such that $a^x =0$ for some "new type of number x", maybe you can win the Fields Prize.
A: You are correct. The equation has no solutions. Not even for complex $x$. 
A: Yes. No solution for the equation $$ 2^x = 0 $$ is defined.
This can be proved using the definition of logarithms.
