Non-integer Powers and Exponents
Date: 01/06/99 at 10:53:41 From: Christina Subject: Taking a number to a power that's not an even number I understand x squared or x cubed, but how do you get x to the 1.9 for instance?
Date: 01/06/99 at 13:11:15 From: Doctor Rob Subject: Re: Taking a number to a power that's not an even number Thanks for writing to Ask Dr. Math! The expression x^(1/n) is defined to be the nth root of x. If you are comfortable with finding the nth root of a number, the rest follows. Then x^(m/n) = [x^(1/n)]^m, so you can compute it as the mth power of an nth root, where m and n are positive whole numbers. Take the exponent you are interested in, such as 1.9, and write it as a fraction 1.9 = 19/10 Then x^(19/10) = [x^(1/10)]^19 In the case of a terminating decimal, you can always use a denominator which is a power of 10. In the case of a repeating decimal, you can always use a denominator that is a power of 10 times one less than another power of 10: 10^a*(10^b-1), for some integers a and b. In the case of an irrational number like sqrt(2) = 1.4142..., you can approximate x^sqrt(2) by x^1, then x^1.4, then x^1.41, then x^1.414, then x^1.4142, and so on, until two successive approximations agree to as many decimal places as you need. Example: Find 5^sqrt(2) to five decimal places of accuracy. 5^1 = 5.0000000000 5^1.4 = 5^(14/10) = [5^(1/10)]^14 = 1.7461894309^14 = 9.5182696936 5^1.41 = 5^1.4*5^(1/100) = 9.5182696936*1.0162245913 = 9.6726997289 5^1.414 = 5^1.41*5^(4/1000) = 9.67269973*1.006458519 = 9.7351710392 5^1.4142 = 9.7383051743 5^1.41421 = 9.7384619075 5^1.414213 = 9.7385089280 5^1.4142135 = 9.7385167647 5^1.41421356 = 9.7385177051 5^1.414213562 = 9.7385177365 So to five decimal places, 5^sqrt(2) = 9.73852. There is another way using logarithms and antilogarithms which can cut down the amount of numerical computation, but requires knowledge of math through calculus to explain in detail. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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