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### Finding a Decimal Exponent

```
Date: 03/21/98 at 15:47:15
From: Tyler
Subject: Decimal exponent

How do you find 32^0.4 without using a calculator? I know it's 4, but
I can't seem to prove why it is that. I've noticed somewhat of a
pattern, not so sure if it is linked. Since 32^0.4 is the same as 2.5
(exponent) radical 32 and I got 4, I took 4^2.5 and got 32. Then I
tried 5^2.5 and got an irrational number. I tried numbers starting
from one and up. I got irrational numbers until 9, 16, and 25, which
are obviously square roots. What's the connection?
```

```
Date: 03/21/98 at 19:50:58
From: Doctor Sam
Subject: Re: Decimal exponent

Tyler,

Decimal exponents can often be evaluated by rewriting the exponent as
a fraction.

0.4 = 4/10 = 2/5, so the problem is 32^0.4 = 32^(2/5). This can be
evaluated in several ways and you figured one of them out. Rewrite
32 = 2^5 so that the problem becomes [2^5]^(2/5).

A basic property of exponents is that [x^a]^b = x^(ab) so
[2^5]^(2/5) = [2]^[(5)(2/5)] = 2^2 = 4.

Another way to get the same result is to interpret the power 2/5.
In general x^(a/b) = [x^a]^(1/b) or  [x^(1/b)]^a. In this case
[32]^(2/5)= [32^(1/5)]^2.  32^(1/5) is the fifth root of 32, which is
2, and once again 2^2 = 4.

The pattern you observed is that only square numbers: 4,9,16,...,x^2
give whole numbers when raised to the 2.5 power. The reason for this
is the law of exponents I mentioned above: [x^a]^b = x^(ab).

If you have a perfect square number x^2 and raise it to the 2.5 power
you are computing: [x^2]^2.5 = x^5. If x is a whole number, then so
is x^5. If the number is not a perfect square (like 5,6,7, or 8) then
when you raise it to the power 2.5 = 5/2 you are computing
N^(5/2) = sqrt(N^5), which will be irrational.

I hope that helps.

Doctor Sam,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Exponents

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