Negative Fraction Exponents
Date: 01/27/99 at 09:39:26 From: Khalida Subject: Pre-calculus Exponents 2n^(1/3) - 4n^(-2/3) ------------------- 2n^(-2/3) I am having trouble with these kinds of problems. If you could help me simplify these exponents and also radical exponents, I would be very thankful. Khalida
Date: 01/27/99 at 12:55:13 From: Doctor Rick Subject: Re: Pre-calculus Exponents Hi, Khalida. I would start by breaking this into a difference of two fractions (by applying the distributive principle). 1/3 -2/3 2n 4n ------ - ------- -2/3 -2/3 2n 2n Now, you can see right off that in each fraction, the numerator and denominator are both divisible by 2. So divide numerators and denominators by 2: 1/3 -2/3 n 2n ----- - ------- -2/3 -2/3 n n What's left is the part you are most concerned about, I'm sure. Here are two rules you can use: x^n / x^m = x^(n-m) 1 / x^n = x^(-n) Use the second rule to convert each fraction (division) into a product: 1/3 2/3 -2/3 2/3 n * n - 2n * n Then use the first rule to combine the powers of n: (1/3 + 2/3) (-2/3 + 2/3) n - 2n Do the additions: 1 0 n - 2n which we can write as n - 2 Wow, that's a lot simpler! You can simplify expressions with radicals in the same way, if you first rewrite each radical as a fractional power, for instance, sqrt(x) = x^(1/2). If you need more help with radicals, send an example, and we'll work through it. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.