audio read-through Practice With $\,\displaystyle x^{-p} = \frac{1}{x^p}$

Recall that $\displaystyle\,x^{-1} = \frac 1x\,.$ That is, $\displaystyle\,x^{-1}\,$ is the reciprocal of $\,x\,.$

It follows, using the exponent laws, that:

$$\cssId{s8}{x^{-p} = (x^p)^{-1} = \frac{1}{x^p}}$$

That is, $\,x^{-p}\,$ is the reciprocal of $\,x^p\,.$

Continuing, it's convenient to notice that expressions of the form $\,x^m\,$ can be moved from numerator to denominator, or from denominator to numerator, just by changing the sign of the exponent.

For example:

$\displaystyle \cssId{s17}{\frac{1}{x^{-3}}} \cssId{s18}{= \frac{x^3}{1}} \cssId{s19}{= x^3}\,$:  exponent was negative in denominator; is positive in numerator

$\displaystyle \cssId{s22}{\frac{1}{x^{3}}} \cssId{s23}{= \frac{x^{-3}}{1}} \cssId{s24}{= x^{-3}}\,$:  exponent was positive in denominator; is negative in numerator

$\displaystyle \cssId{s27}{x^3} \cssId{s28}{= \frac{x^3}{1}} \cssId{s29}{= \frac{1}{x^{-3}}}\,$:  exponent was positive in numerator; is negative in denominator

$\displaystyle \cssId{s32}{x^{-3}} \cssId{s33}{= \frac{x^{-3}}{1}} \cssId{s34}{= \frac{1}{x^3}}\,$:  exponent was negative in numerator; is positive in denominator

All the exponent laws are stated below, for completeness.


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.


$x^mx^n = x^{m+n}$ 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


$\displaystyle \frac{1}{x^3} = x^p\,$ where $\,p = \text{?}$
Answer: $p = -3$
$\displaystyle \frac{1}{x^{-2}} = x^p\,$ where $\,p = \text{?}$
Answer: $p = 2$