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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 $x=-\infty$?

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@user134 what is the problem with upvoting this question? –  user42912 May 25 '13 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 '13 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. –  1015 May 25 '13 at 2:22
@julien I got your point. Thanks –  rst May 25 '13 at 2:52
Do you know the definition of $2^x$? –  Jack May 25 '13 at 13:07

6 Answers 6

up vote 19 down vote accepted

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 $0$.

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this means $2^x=0$ is possible for x = -∞ . –  rst May 25 '13 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. –  dfeuer May 25 '13 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.

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$+ \cos^2 x + \sin^2 x$ (I hope this answer gets more exposure!) –  amWhy May 25 '13 at 4:16
$+i^{4}$ Does the extension to $\mathbb{R}^*$ work out, or do bad things happen? –  Lucas May 25 '13 at 4:21
@Lucas, what do you consider bad? –  dfeuer May 25 '13 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. –  Lucas May 25 '13 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. –  Lucas May 25 '13 at 6:05

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

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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 $0$:


The graph is not a proof this function is never $0$, it's just to illustrate what others have said in another answers graphically.

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@PeterTamaroff I'm sorry I didn't understand, what did you mean? –  user42912 May 25 '13 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 Tamaroff May 25 '13 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. –  user42912 May 25 '13 at 2:59
@PeterTamaroff I edited the answer, is it better now? –  user42912 May 25 '13 at 3:01

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

This can be proved using the definition of logarithms.

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"This can be proved using the definition of logarithms." How do you define logarithms in the first place? –  Pedro Tamaroff May 25 '13 at 2:34

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.

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Well, $2^2 = 0$ in $\mathbb{Z}/4\mathbb{Z}$. Can I have my Fields Medal now? –  Tobias Kildetoft May 25 '13 at 2:25
Such a number already exists in the extended reals. –  alex.jordan May 25 '13 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. –  1015 May 25 '13 at 2:33
Well, Galois deserves one. –  hyg17 May 25 '13 at 2:37
@alex.jordan you mean $2^x=0$ if x= -∞ in extended real [-∞,∞] –  rst May 25 '13 at 2:47

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