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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 $x=-\infty$?

rst
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6 Answers6

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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 '13 at 04:55
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    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 05:19
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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.

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

learner
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Anthony Deluca
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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 $0$:

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.

user42912
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  • @PeterTamaroff I'm sorry I didn't understand, what did you mean? – user42912 May 25 '13 at 02:29
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    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 '13 at 02:32
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    @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 02:59
  • @PeterTamaroff I edited the answer, is it better now? – user42912 May 25 '13 at 03:01
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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.

hyg17
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Yes. No solution for the equation $$ 2^x = 0 $$ is defined.

This can be proved using the definition of logarithms.

hjpotter92
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    "This can be proved using the definition of logarithms." How do you define logarithms in the first place? – Pedro May 25 '13 at 02:34