# Math: two exponent questions
**URL:** https://boards.straightdope.com/t/math-two-exponent-questions/40358
**Category:** Factual Questions
**Created:** [November 8, 2000, 8:18pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358 "2000-11-08T20:18:12Z")
**Posts on this page:** 20
**Page:** 1
### Author:  [@zev\_steinhardt](https://boards.straightdope.com/u/zev_steinhardt)
#### Post date: [November 8, 2000, 8:18pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/1 "2000-11-08T20:18:12Z")
Two quick questions concerning exponents:
1. We all know the following to be true:
[ul]
[li]2[sup]1[/sup] = 2[/li][li]2[sup]2[/sup] = 4[/li][li]2[sup]3[/sup] = 8[/li][li]2[sup]4[/sup] = 16[/li][li]2[sup]5[/sup] = 32[/li][/ul]
What is 2[sup]4.5[/sup] and how is that calculated?
1. Why is X[sup]0[/sup] = 1 where X = any number?
Zev Steinhardt
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### Author:  [@AWB](https://boards.straightdope.com/u/AWB)
#### Post date: [November 8, 2000, 8:30pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/2 "2000-11-08T20:30:05Z")
Quick answer: 2[sup]4.5[/sup]
= 2[sup]9/2[/sup]
= (2[sup]9[/sup])[sup]1/2[/sup]
= sqrt(2[sup]9[/sup])
= sqrt(512)
= 22.6274
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### Author:  [@zev\_steinhardt](https://boards.straightdope.com/u/zev_steinhardt)
#### Post date: [November 8, 2000, 8:33pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/3 "2000-11-08T20:33:50Z")
> [@](#):
>
> \*Originally posted by AWB \*
> \*\*Quick answer: 2[sup]4.5[/sup]
> = 2[sup]9/2[/sup]
> = (2[sup]9[/sup])[sup]1/2[/sup]
> = sqrt(2[sup]9[/sup])
> = sqrt(512)
> = 22.6274 \*\*
Thank you **AWB**. I never would have thought to figure it out that way. Do you know the answer to my second question?
Zev Steinhardt
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### Author:  [@CalMeacham](https://boards.straightdope.com/u/CalMeacham)
#### Post date: [November 8, 2000, 8:35pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/4 "2000-11-08T20:35:45Z")
Zev:
You ought to be able to find the answers to your questions in any good advanced algebra text, or an introductory calculus book. I know it’s in my copy of George B. Thomas’ “Calculus”.
For the first question, recall that squaring x to the n is the same as x to the 2 n. In other words, (x^n)^2 = x^2n. Taking the square root ought thus to be the same as dividing the exponent by 2: SQRT(x^n) = x^(n/2) In your case x^4.5 = SQRT(x^9)
Since (x^n)/(x^m) = x^(n-m), you can see that 1 = (x^n)/(x^n) = x^0
This should hold true for any x, with the possible exception of x = 0. However, you find that a lot of sequences and series can be written succinctly if you define 0^0 = 1. There is, in fact, a reference in a footnote in Thomas’ book about this.
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### Author:  [@KneadToKnow](https://boards.straightdope.com/u/KneadToKnow)
#### Post date: [November 8, 2000, 8:37pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/5 "2000-11-08T20:37:54Z")
2[sup]2[/sup]=2_2_1
2[sup]1[/sup]=2\*1
2[sup]0[/sup]=1
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### Author:  [@Cantrip](https://boards.straightdope.com/u/Cantrip)
#### Post date: [November 8, 2000, 8:38pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/6 "2000-11-08T20:38:37Z")
> [@](#):
>
> \*Originally posted by zev\_steinhardt \*
> \*\*Two quick questions concerning exponents:
>
> 1. We all know the following to be true:
> [ul]
> [li]2[sup]1[/sup] = 2[/li][li]2[sup]2[/sup] = 4[/li][li]2[sup]3[/sup] = 8[/li][li]2[sup]4[/sup] = 16[/li][li]2[sup]5[/sup] = 32[/li][/ul]
> What is 2[sup]4.5[/sup] and how is that calculated?
> Zev Steinhardt \*\*
Well, we know that x[sup]n[/sup]\*x[sup]m[/sup]=x[sup]n+m[/sup]. For example, 2[sup]2[/sup]_2[sup]3[/sup]=2[sup]5[/sup]=32, which also = 4_8.
So, 2[sup]4.5[/sup]=2[sup]4[/sup]_2[sup]1/2[/sup], which equals 16_square root of 2.
I think the other one is definitional, although I’m sure there’s a mathematical explanation. Try graphing it. (I always hated when my teachers told me that.) 🙂
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### Author:  [@AWB](https://boards.straightdope.com/u/AWB)
#### Post date: [November 8, 2000, 8:39pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/7 "2000-11-08T20:39:16Z")
_1. Fractional exponents_
If the exponent on a symbol _x_ is a fraction _p/q_, then _x[sup]p/q[/sup]_ is defined as _(x[sup]1/q[/sup])[sup]p[/sup]_, where _x[sup]1/q[/sup]_ is the positive _q_th root of _x_ if _x_ is positive, and the (negative) _q_th root if _x_ is negative and _q_ is odd. It follows that _x[sup]p/q[/sup] = (x[sup]p[/sup])[sup]1/q[/sup]_.
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### Author:  [@AWB](https://boards.straightdope.com/u/AWB)
#### Post date: [November 8, 2000, 8:48pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/8 "2000-11-08T20:48:28Z")
1. Zero Exponent
According to the 2nd law of exponents:
_a[sup]m[/sup]/a[sup]n[/sup] = a[sup]m-n[/sup]_
If both _m_ and _n_ are equal, say to _p_, then:
_a[sup]p[/sup]/a[sup]p[/sup] = a[sup]p-p[/sup]_
The left side of this equation is 1, since a number divided by itself is 1. The right side becomes _a[sup]0[/sup]_. So,
_a[sup]0[/sup] = 1_
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### Author:  [@AWB](https://boards.straightdope.com/u/AWB)
#### Post date: [November 8, 2000, 8:56pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/9 "2000-11-08T20:56:02Z")
Another way to figure non-integer exponential expressions:
Say you have an expresstion _a[sup]y[/sup]_, where _y_ is a non-integer expression (and perhaps not an easy decimal expression such as the OP’s 4.5).
Let V = _a[sup]y[/sup]_
_ln_ V = _ln a[sup]y[/sup]_
_ln_ V = _y ln a_
_e[sup]ln_ V[/sup] = _e[sup]y ln a[/sup]_
V = _e[sup]y ln a[/sup]_
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### Author:  [@sailor](https://boards.straightdope.com/u/sailor)
#### Post date: [November 8, 2000, 11:13pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/10 "2000-11-08T23:13:20Z")
If the limit of X^Y as Y approaches 0 from both sides is 1 then I think it makes sense to define X^0=1 (unless the electoral college has a better idea).
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### Author:  [@scr4](https://boards.straightdope.com/u/scr4)
#### Post date: [November 9, 2000, 1:00am UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/11 "2000-11-09T01:00:57Z")
> [@](#):
>
> [ul]
> [li]2[sup]5[/sup] = 32[/li][li]2[sup]4[/sup] = 16[/li][li]2[sup]3[/sup] = 8[/li][li]2[sup]2[/sup] = 4[/li][li]2[sup]1[/sup] = 2[/li][/ul]
> \*\*
[ul]
[li]2[sup]0[/sup] = 1[/li][li]2[sup]-1[/sup] = 1/2[sup]1[/sup] = 1/2[/li][li]2[sup]-2[/sup] = 1/2[sup]2[/sup] = 1/4[/li][li]2[sup]-3[/sup] = 1/2[sup]3[/sup] = 1/8[/li][/ul]
This is not a proof, but I think it helps to show that 2[sup]0[/sup]=1 is natural, or at least makes sense.
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### Author:  [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)
#### Post date: [November 9, 2000, 3:34pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/12 "2000-11-09T15:34:28Z")
Just keep in mind that X^0 = 1 only as long as X is not zero. For example, in AWB’s last two posts, if _a_ = 0, he either divides by zero, or takes the log of zero, both of which are undefined.
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### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [November 9, 2000, 6:16pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/13 "2000-11-09T18:16:29Z")
0[sup]0[/sup] depends on the context in which you’re asking. In most situations where it comes up, it makes things simpler to define it as being 1. For instance, lim(x --\> +0) x[sup]x[/sup] = 1 (I’m not sure about the limit from the left). However, there’s no consistent way to define it for all cases.
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### Author:  [@Joe\_Cool](https://boards.straightdope.com/u/Joe_Cool)
#### Post date: [November 9, 2000, 7:57pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/14 "2000-11-09T19:57:47Z")
> [@](#):
>
> \*Originally posted by AWB \*
> \*\*Another way to figure non-integer exponential expressions:
>
> Say you have an expresstion _a[sup]y[/sup]_, where _y_ is a non-integer expression (and perhaps not an easy decimal expression such as the OP’s 4.5).
>
> Let V = _a[sup]y[/sup]_
> _ln_ V = _ln a[sup]y[/sup]_
> _ln_ V = _y ln a_
> _e[sup]ln_ V[/sup] = _e[sup]y ln a[/sup]_
> V = _e[sup]y ln a[/sup]_
>
> \*\*
AWB:
The problem with that explanation is that you’re still left with raising a number (e) to a fractional power (a trancendental one, no less!). So basically you’re stuck with a harder version of the same problem you started out with.
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### Author:  [@John\_Kentzel-Griffin](https://boards.straightdope.com/u/John_Kentzel-Griffin)
#### Post date: [November 10, 2000, 4:00pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/15 "2000-11-10T16:00:23Z")
**Joe\_Cool** makes a excellent point. Let me see if I can simplify. (Is simplify the word I’m looking for?)
Exponentiation to a positive power is a compact way to represent repeated multiplication. Using this as a definition, only positive integral powers are defined. We notice that exponentiation follows the following rules:[list=1][li]a[sup]1[/sup] = a[_]a[sup]m[/sup] \* a[sup]n[/sup] = a[sup]m+n[/sup]\*[sup]m[/sup] = a[sup]m_n[/sup]a[sup]m[/sup]/a[sup]n[/sup] = a[sup]m-n[/sup] (for m \> n, a not 0)[/list=1]The proofs are left as exercises for the reader.[/li]
We would like to extend the definition of exponentiation to allow for exponents that are not positive integers while still following the above rules. We can define exponentiation as an operation that follows the above rules (removing the restriction m\>n in rule 4). It is convenient to require the base to be a positive number when using an exponent that is not a positive number. By applying rule 4 with m=n we have:
a[sup]m[/sup]/a[sup]m[/sup] = a[sup]m-m[/sup]
1 = a[sup]0[/sup]
a[sup]1/2[/sup] = sqrt( (a[sup]1/2[/sup])[sup]2[/sup])
= sqrt(a[sup]1/2 \* 2[/sup])
= sqrt(a)
Using this result;
2[sup]4.5[/sup] = 2[sup]9\*1/2[/sup]
= sqrt(2[sup]9[/sup])
= sqrt(2[sup]8[/sup]\*2)
= 2[sup]4[/sup]_sqrt(2)
= 16_sqrt(2)
This answers the OP.
The astute reader will notice that this definition of exponentiation (an operation that obeys rules 1-4) only defines rational powers. Another rule must be added to extend to irrational powers.
I’ve never used Preview Reply more times than for this post.
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### Author:  [@egkelly](https://boards.straightdope.com/u/egkelly)
#### Post date: [November 10, 2000, 5:19pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/16 "2000-11-10T17:19:18Z")
What is i (square root of -1) to the ith power?
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### Author:  [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)
#### Post date: [November 10, 2000, 5:56pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/17 "2000-11-10T17:56:12Z")
> [@](#):
>
> What is i (square root of -1) to the ith power?
e^(-pi/2) = 0.20787957635076
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### Author:  [@John\_Kentzel-Griffin](https://boards.straightdope.com/u/John_Kentzel-Griffin)
#### Post date: [November 10, 2000, 6:05pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/18 "2000-11-10T18:05:26Z")
\*_ZenBeam_
I’m not sure how one would define a non-positive real raised to a non-integral exponent, let alone an imaginary raised to an imaginary. A complex base may be raised to a positive integral (real) exponent, and a positive (real) number may be raised to a complex exponent.
How does one define a complex raised to a complex? I’m not saying you are incorrect, I just want to know where your answer came from since you didn’t show your work. 🙂
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### Author:  [@panamajack](https://boards.straightdope.com/u/panamajack)
#### Post date: [November 10, 2000, 6:16pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/19 "2000-11-10T18:16:08Z")
I think the point of AWB’s analysis is that natural logs have been tabulated and are thus more easily computable than arbitrary roots (which is what you’d have to work out otherwise (see AWB’s first post)). This is what calculators & log tables are for; if you want more precise answers, do the series.
And ZenBeam, I’m not sure on your answer; this is what I get :
```auto
Find X = j[sup]j[/sup]
log[sub]j[/sub]X = j (rewrite)
ln X / ln j = j (change of base)
ln X = j ln j (multiply by ln j)
ln X = j (pi/2) (ln j = pi/2 by Euler's Formula)
X = exp(j (pi/2)) (take exp() of both sides)
```
Am I missing something?
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### Author:  [@John\_Kentzel-Griffin](https://boards.straightdope.com/u/John_Kentzel-Griffin)
#### Post date: [November 10, 2000, 6:36pm UTC](https://boards.straightdope.com/t/math-two-exponent-questions/40358/20 "2000-11-10T18:36:27Z")
> [@](#):
>
> \*Posted by \ ***panamajack**
> ln j = pi/2 by Euler’s Formula
I don’t know Euler’s Formula off the top of my head, but the above implies (by raising both sides to e):
_i_ = e[sup]pi/2[/sup]
😕
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