﻿ Multi-step Exponent Law Practice
MULTI-STEP EXPONENT LAW PRACTICE

In this exercise you will practice with the exponent laws, all mixed-up.

These problems require the application of more than one exponent law.
For simpler problems, see One-Step Exponent Law Practice.

EXPONENT LAWS
Let $\,x\,$, $\,y\,$, $\,m\,$, and $\,n\,$ be real numbers, with the following exceptions:
• a base and exponent cannot simultaneously be zero (since $\,0^0\,$ is undefined);
• division by zero is not allowed;
• for non-integer exponents (like $\,\frac12\,$ or $\,0.4\,$), assume that bases are positive.
Then,
 $x^mx^n = x^{m+n}$ Verbalize: same base, things multiplied, add the exponents $\displaystyle \frac{x^m}{x^n} = x^{m-n}$ Verbalize: same base, things divided, subtract the exponents $(x^m)^n = x^{mn}$ Verbalize: something to a power, to a power; multiply the exponents $(xy)^m = x^my^m$ Verbalize: product to a power; each factor gets raised to the power $\displaystyle \left(\frac{x}{y}\right)^m = \frac{x^m}{y^m}$ Verbalize: fraction to a power; both numerator and denominator get raised to the power
EXAMPLES:
 $\displaystyle\left(\frac{1}{x^2}\right)^3 = (x^{-2})^3 = x^{-2\,\cdot\, 3} = x^{-6} = x^p$ where $\,p = -6$ $\displaystyle \left(\frac{x^2}{x^3}\right)^5 = (x^{2-3})^5 = (x^{-1})^5 = x^{-1\,\cdot\, 5} = x^{-5} = x^p$ where $p = -5$ $(x^2x^4)^{-1} = (x^{2+4})^{-1} = (x^6)^{-1} = x^{6\,\cdot\, -1} = x^{-6} = x^p$ where $\,p = -6$ $\displaystyle \frac{x^2x^{-3}}{x^5} = \frac{x^{2 + (-3)}}{x^5} = \frac{x^{-1}}{x^5} = x^{-1-5} = x^{-6} = x^p$ where $\,p = -6$ $\displaystyle \frac{x^2}{x^3x^4} = \frac{x^2}{x^{3+4}} = \frac{x^2}{x^7} = x^{2-7} = x^{-5} = x^p$ where $\,p = -5$ $\displaystyle \frac{(x^2)^3}{(x^{-1})^4} = \frac{x^{2\,\cdot\,3}}{x^{-1\,\cdot\,4}} = \frac{x^6}{x^{-4}} = x^{6-(-4)} = x^{10} = x^p$ where $\,p = 10$
Master the ideas from this section