Finding a Decimal ExponentDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/