Simplify Expressions with Exponents -

Simplify a variety of algebraic expressions using the laws of exponents.

laws of exponents

Example 1

Simplify $3x^4\cdot 5 x^2$

It is important to remember that the exponent applies only to what it is written beside. The exponent $4$ applies only to the $x$ . Written out in full we have:

$\begin{align*}3x^4\cdot 5 x^2&=3\cdot x\cdot x\cdot x \cdot x \cdot 5 \cdot x \cdot x\\[10 pt]&=3\cdot 5\cdot x\cdot x \cdot x \cdot x \cdot x \cdot x\\[10 pt]&=15x^6\end{align}$

In short, $3x^4\cdot 5 x^2=15x^6$

Example 2

Simplify $5(x^4y)^3$

The exponent 3 is applied to each factor in the bracket:

$\begin{align*}5(x^4y)^3&=5(x^4y)(x^4y)(x^4y)\\[10 pt]&=5\cdot x^4\cdot x^4\cdot x^4\cdot y\cdot y\cdot y\\[10 pt]&=5x^{12}y^3\end{align}$

In short, $5(x^4y)^3=5x^{12}y^3$

Example 3

Simplify $4^{-3}$

$4$ raised to the exponent negative $3$ means to divide by $4$ three times. We ‘take the reciprocal‘. The result here is a positive number.

$\begin{align*}4^{-3}&=\\[10 pt]&=\dfrac{1}{4^3}\\[10 pt]&=\dfrac{1}{64}\end{align}$

In short, $4^{-3}=\dfrac{1}{64}$

Example 4

Simplify $\dfrac{1}{4^{-3}}$

Let’s see what the result is of a negative exponent on the denominator in general:

Simplify $\dfrac{1}{a^{-m}}$

$\begin{align*}\dfrac{1}{a^{-m}}&=\dfrac{1}{\frac{1}{a^m}}\\[10 pt]&=1\div\dfrac{1}{a^m}\\[10 pt]&=1 \times \dfrac{a^m}{1}\\[10 pt]&=a^m\end{align}$

In short, $\dfrac{1}{a^{-m}}=a^m$

Applying this new law, we have $a=4$ and $m=3$ .

$\dfrac{1}{4^{-3}}=4^3=64$

Example 5

Simplify $\dfrac{5x^7y^4}{15x^{-3}y^6}$

It is helpful to gather like factors as follows:

$\begin{align*}\dfrac{5x^7y^4}{15x^{-3}y^6}&=\left(\dfrac{5}{15}\right)\left(\dfrac{x^7}{x^{-3}}\right)\left(\dfrac{y^4}{y^6}\right)\\[10 pt]&=\left(\dfrac{1}{3}\right)\left(x^{10}\right)\left(y^{-2}\right)\\[10 pt]&=\dfrac{x^{10}}{3y^2}\end{align}$

Example 6

Simplify $\left(\dfrac{5x^7y^4}{15x^{-3}y^6}\right)^{-2}$

Let’s make a law for a fraction raised to a negative exponent:

Simplify $\left(\dfrac{a}{b}\right)^{-m}$

$\begin{align*}\left(\dfrac{a}{b}\right)^{-m}&=\dfrac{a^{-m}}{b^{-m}}\\[10 pt]&=a^{-m}\cdot \dfrac{1}{b^{-m}}\\[10 pt]&=\dfrac{1}{a^m}\cdot b^m\\[10 pt]&=\left(\dfrac{b}{a}\right)^m\end{align}$

In short, $\left(\dfrac{a}{b}\right)^{-m}=\left(\dfrac{b}{a}\right)^m$

Apply this new law to example 6 we have

$\begin{align*}\left(\dfrac{5x^7y^4}{15x^{-3}y^6}\right)^{-2}&=\left(\dfrac{15x^{-3}y^6}{5x^{7}y^4}\right)^{2}\\[10 pt]&=\left(\left(\dfrac{15}{5}\right)\left(\dfrac{x^{-3}}{x^7}\right)\left(\dfrac{y^6}{y^4}\right)\right)^2\\[10 pt]&=\left(3x^{-10}y^2\right)^2\\[10 pt]&=\dfrac{9y^4}{x^{20}}\end{align}$

Practice

Try the ten questions at the end of this mathisfun page.

applet link

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