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Hi all I know that $5^1 = 5$, $5^2 = 25$, $5^3 = 125$.

But why is $5^{1.5} = 11.180339887498949$ ?

How did we get the number $11.180339887498949$ ?

Brian M. Scott
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Pacerier
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  • Have you heard of these logarithms, if not you can read this wikipedia article: http://en.wikipedia.org/wiki/Logarithm –  Jan 12 '12 at 09:27
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    $5^{1.5}=5^{1+0.5}=5\times5^{0.5}=5\sqrt{5}$. – J. M. ain't a mathematician Jan 12 '12 at 09:29
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    We know that, for example, $(5^2)^3 = 5^{2\times 3}$. So, whatever $5^{1.5}$ might be, we'd like it to satisfy $(5^{1.5})^2 = 5^{1.5\times 2} = 5^3 = 125$. So $5^{1.5}$ ought to be the square root of 125. –  Jan 12 '12 at 09:30
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    To avoid any confusion about irrationals, consider instead $4^{1.5} = 8$. – Dan Brumleve Jan 12 '12 at 09:31
  • @Rahul Narain, This is the answer. Why did you make it a comment? – Lieven Jan 12 '12 at 10:03
  • @RahulNarain But this is like a chicken-egg problem. So I understand that 5^1.5 = 5 × sqrt(5). But how do you solve for sqrt(5) in the first place? – Pacerier Jan 12 '12 at 11:01
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    @Lieven, because I felt like the answer this question really ought to get is something long, comprehensive, and impressive, like [André Nicolas' explanation of irrational exponents](http://math.stackexchange.com/a/55078/856), and I didn't have time to write one. –  Jan 12 '12 at 11:05
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    @Pacerier, how do we know that $\sqrt 5 = 2.23606...$? Well, 2 is too small because its square is 4, but 3 is too big because its square is 9, similarly 2.2 is too small, 2.3 is too big, 2.23 is too small, 2.24 is too big, and so on... Of course, there are [more efficient methods](http://en.wikipedia.org/wiki/Nth_root#Computing_principal_roots). But this is something [you could have invented](http://www.google.com/search?q=you+could+have+invented) yourself, and I find such explanations to be the most edifying. –  Jan 12 '12 at 11:14
  • @RahulNarain do you mean to say that when I key sqrt(5) in my calculator, it's using that algorithm (which you linked) to solve for sqrt(5) ? – Pacerier Jan 12 '12 at 11:26
  • Well, gee, how am I supposed to know how your calculator does it? I don't even know how *my* calculator does it! ...More seriously, I think kahen and I have answered the question you asked in your main post. If you want to know specifically about the algorithms typically used in calculators, you should edit that into your question (or ask a new one) and I'm sure someone with the requisite expertise will come along and answer. (I'd wager a calculator with an $x^y$ function probably computes it [via $\exp(y\log x)$](http://math.stackexchange.com/q/55068/856) instead, but I'm just speculating.) –  Jan 12 '12 at 11:56
  • @RahulNarain ok thanks for the help =D – Pacerier Jan 12 '12 at 12:20

2 Answers2

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If $a,b$ are positive integers and $x > 0$, then $x^{a/b}$ is defined as $\sqrt[b]{x^a}$, where the $b^\text{th}$ root of $y>0$ is the unique positive real number $r$ such that $r^b = y$. So $5^{1.5}$ = $\sqrt{5^3}$, i.e. it's the number which squared is $125$.

This leaves open many questions such as "why do $n^\text{th}$ roots exist?" and "what about $x^\alpha$ where $\alpha$ is irrational?"

There are no easy answers to those questions which don't involve a first course in real analysis. If you haven't done university level real analysis, you kind of have to take on faith that exponentiation works and obeys the rules given. I know that's a disappointing answer, but it's the only decent answer I have.

kahen
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  • But this is like a chicken-egg problem. So I understand that 5^1.5 = 5 × sqrt(5). But how do you solve for sqrt(5) in the first place? – Pacerier Jan 12 '12 at 11:00
  • Like I said: there is no easy answer. You either construct a [sequence](http://en.wikipedia.org/wiki/Sequence_%28mathematics%29) converging to $\sqrt{5}$ (think of succesively better approximate solutions to the equation $x^2 = 5$) or you realize it as the [supremum](http://en.wikipedia.org/wiki/Supremum) of e.g. the set $\lbrace x \in \mathbb R \;|\; x^2 \leq 5\}$. Here's a note that explains it, but I think it's going to be way over the level you're at: [PDF](http://arxiv.org/pdf/0805.3303). It was not without reason I wrote that you have to take a lot of things on faith about exponentiation – kahen Jan 12 '12 at 11:10
  • Thanks for the link to that PDF Kahen! – Samuel Reid Jan 13 '12 at 00:42
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$$ 2\cdot2\cdot2=8, $$ so multiplying by $2$ three times is the same as multiplying by $8$,

and multiplying by $8$ one-third of one time is the same as multiplying by $2$.

Michael Hardy
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