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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?

ehuang
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3 Answers3

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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.

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    The source of my confusion is: $8^{1/2} = \sqrt{8}$ or $-\sqrt{8}$. – ehuang Aug 03 '13 at 08:05
  • @EricHuang: Usually, $\log(8)$ is taken to be a real number. If so, then $8^{1/2}=e^{\frac12\large\log(8)}$ is a positive real. – robjohn Aug 03 '13 at 10:42
3

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.

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

enter image description here

Daniel Franke
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  • What's the plot for? – Pedro Aug 03 '13 at 03:30
  • I voted this down: A plot may suggest an answer in mathematics, but it is not an answer in itself. For example, from your response, how can we conlclude that $8^x \ge 0$ when $|x| > 3$? – Stephen Herschkorn Aug 03 '13 at 04:00
  • @StephenHerschkorn A reasonable answer is given right in question: while it is perhaps in need of rigor, there's no reason that such rigor cannot be easily be added to the OP's argument. The OP seems confused, though, thinking his conclusion is false. The plot is intended as an aid to help the OP identify the source of his confusion. – Daniel Franke Aug 03 '13 at 04:06