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