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

When you're done practicing, move on to:
Practice with Radicals

 
 
Simplify:
    
(an even number, please)