# 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 $x=-\infty$?

• @user134 what is the problem with upvoting this question? May 25, 2013 at 2:15
• @user42912 there's no problem at all. You're free to upvote anything you want, and that's why I deleted my comment. :)
– ՃՃՃ
May 25, 2013 at 2:18
• The equation is not wrong per se, it is just that $\{x\in\mathbb{R}\;;\;2^x=0\}=\emptyset$. In $\mathbb{C}$ also, btw. May 25, 2013 at 2:22
• @julien I got your point. Thanks
– rst
May 25, 2013 at 2:52
• Do you know the definition of $2^x$?
– user9464
May 25, 2013 at 13:07

## 6 Answers

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 $0$.

• this means $2^x=0$ is possible for x = -∞ .
– rst
May 25, 2013 at 4:55
• Correct, but you have to be careful about context. If you're looking for a real or complex $x$ such that $2^x = 0$, you won't find one. And as user17762 implies, you have to be careful to actually define exponentiation on the extended real line before you can give an answer in that context. May 25, 2013 at 5:19

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.

• $+i^{4}$ Does the extension to $\mathbb{R}^*$ work out, or do bad things happen? May 25, 2013 at 4:21
• @Lucas, what do you consider bad? May 25, 2013 at 5:22
• @dfeuer Volcanoes erupting in populated areas. Daytime television at night. UKIP forming a majority government. Power failures. Jelly not setting properly. The coming Gozer. Rising sea levels. May 25, 2013 at 6:00
• @dfeuer Well, $\mathbb{R}^*$ with $+,-,\times,/$has some issues ($\infty + (-\infty)$ etc). Just wondering if defining exponential and logarithmic add anything extra to the list of things one has to avoid by leaving something undefined. May 25, 2013 at 6:05
• @Lucas: The extension to $\Bbb R^*$ works just fine, that's why I said "...extend $2^x$ continuously to...".
– Pedro
May 25, 2013 at 13:01

You are correct. The equation has no solutions. Not even for complex $x$.

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.

• @PeterTamaroff I'm sorry I didn't understand, what did you mean? May 25, 2013 at 2:29
• You're only exhibiting the graph for $[-2,2]$. How do we know if $2^x$ is not $0$ elsewhere? My point, in the first place, was that in order to plot $2^x$, you need to have certain information about it. But, in the general case, a graph for $[-2,2]$ is no proof at all.
– Pedro
May 25, 2013 at 2:32
• @PeterTamaroff yes, you're right it's not a proof at all, I plot the graph just to illustrate what I said about the function. May 25, 2013 at 2:59
• @PeterTamaroff I edited the answer, is it better now? May 25, 2013 at 3:01

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.

• Well, $2^2 = 0$ in $\mathbb{Z}/4\mathbb{Z}$. Can I have my Fields Medal now? May 25, 2013 at 2:25
• Such a number already exists in the extended reals. May 25, 2013 at 2:25
• @TobiasKildetoft I'll be happy to write a recommendation letter. Unfortunately, my influence in these circles is equal to the rhs for your astonishing equation. May 25, 2013 at 2:33
• Well, Galois deserves one. May 25, 2013 at 2:37
• @alex.jordan you mean $2^x=0$ if x= -∞ in extended real [-∞,∞]
– rst
May 25, 2013 at 2:47

Yes. No solution for the equation $$2^x = 0$$ is defined.

This can be proved using the definition of logarithms.

• "This can be proved using the definition of logarithms." How do you define logarithms in the first place?
– Pedro
May 25, 2013 at 2:34