Exponents and radicals

In this lesson we are going to dive a little deeper into the world of exponents. If you don’t already have the fundamentals down, check out the pre-algebra lesson on basic exponents.

Multiplying and dividing exponents

When you multiply two exponents of the same variable you simply add the powers to get your new power.

$$\begin{align*} & x^{3}\cdot x^{5}= \\ & x^{3+5}=x^{8} \end{align*}$$

To understand why this works it’s helpful to just write down all the multiplications implied by the exponents.

$$\begin{align*} & x^{3}\cdot x^{5}=(xxx)\cdot(xxxxx)= \\ & xxxxxxxx=x^{8} \end{align*}$$

When you divide two exponents of the same variable you subtract the power of the denominator from the power of the numerator.

$$\frac{x^{7}}{x^{3}}=x^{7-3}=x^{4}$$

Again, expanding the exponents can help you see how this works:

$$\begin{align*} & \frac{x^7}{x^3}= \\[5pt] & \frac{xxxxxxx}{xxx}= \\[5pt] & \frac{xxxx\cancel{xxx}}{\cancel{xxx}}= \\[5pt] & xxxx= \\[5pt] & x^4 \end{align*}$$

When you have several variables in an expression you can apply the division rule to each set of similar variables.

$$\begin{align*} & \frac{12x^{4}y^{5}z}{3x^{3}yz^{3}}=4x^{(4-3)}y^{(5-1)}z^{(1-3)}= \\ & 4xy^{4}z^{-2}=\frac{4xy^{4}}{z^{2}} \end{align*}$$

Power laws

Sometimes we will raise an exponent to another power, like \( (x^{2})^{3} \). The rule here is to multiply the two powers, and it can be shown by expanding the terms.

$$\begin{align*} & (x^{2})^{3}= \\ & x^{2}\cdot x^{2}\cdot x^{2}= \\ & xx\cdot xx\cdot xx= \\ & xxxxxx=x^{6} \\ & (=x^{2\cdot3}) \end{align*}$$

When you have multiple variables all raised to the same power you apply the rule to each of them:

$$\begin{align*} & (x^{2}y^{3})^{4}= \\[0.5em] & x^{2\cdot4}y^{3\cdot4}= \\[0.5em] & x^{8}y^{12} \end{align*}$$

Here’s another example:

$$(\frac{x^{5}}{y^{2}})^{3}=\frac{x^{5\cdot3}}{y^{2\cdot3}}=\frac{x^{15}}{y^{6}}$$

Fractional exponents

We already introduced the square root \( (\sqrt{x^{2}}=x) \), but its possible to have roots of higher powers. For example, \( 4^{3}=64 \) and the ‘cube root’ of 64 is 4: \( \sqrt[3]{64}=4 \). The superscript before the root sign (in this case 3) indicates the power which is being ‘reversed.’ In general, \( \sqrt[n]{x^{n}}=x \).

Roots can also be written as fraction exponents:

$$\begin{align*} & x^{\frac{1}{2}}=\sqrt{x} \\ & x^{\frac{1}{5}}=\sqrt[5]{x} \\ & x^{\frac{1}{n}}=\sqrt[n]{x} \end{align*}$$

We can apply the same multiplication, division and power laws to fractional exponents. For example, multiplication:

$$x^{4}\cdot x^{\frac{1}{3}}=x^{4\frac{1}{3}}=x^{\frac{13}{3}}$$

Division:

$$\frac{x^{2}}{x^{\frac{3}{2}}}=x^{2-\frac{3}{2}}=x^{\frac{1}{2}}$$

Power:

$$(x^{\frac{2}{3}})^{2}=x^{\frac{4}{9}}$$

You can also raise exponents to fractional powers:

$$\begin{align*} & \sqrt[3]{x^{6}y^{2}}=(x^{6}y^{2})^{\frac{1}{3}}= \\ & x^{6\cdot\frac{1}{3}}y^{2\cdot\frac{1}{3}}=x^{2}y^{\frac{2}{3}} \end{align*}$$

Radical expressions

A radical expression is a fraction with radicals (roots) in its denominator. To solve some more difficult problems you will need to be able to ‘rationalize the denominator’ of the expression, which means evicting the radicals. We can do this in certain cases by multiplying by a clever form of \( 1 \). For example say we have this radical expression:

$$\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}$$

We can rationalize the denominator by multiplying the expression by \( \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}} \). Since this is equal to \( 1 \) the multiplication does not change the value of the expression, but it does change the form.

$$\begin{align*} & \frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\cdot\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}= \\[1em] & \frac{\sqrt{x}\sqrt{x}-\sqrt{x}\sqrt{y}}{\sqrt{x}\sqrt{x}-\sqrt{x}\sqrt{y}+\sqrt{x}\sqrt{y}-\sqrt{y}\sqrt{y}}= \\[1em] & \frac{x-\sqrt{xy}}{x-y} \end{align*}$$

The resulting expressions is equivalent to the first for all positive values (you can check this by plugging numbers into each), but it is in a form without radicals in the denominator.

Practice questions

Simplify the following expressions using the different exponent laws

1. \( x^{2}y^{3}\cdot x^{3}y \)
2. \( (xy^{2})^{4}\cdot(x^{2}y)^{3} \)
3. \( \frac{(xy)^{5}}{xy^{2}} \)
4. \( \frac{10xy^{2}z^{3}}{2x^{2}yz} \)
5. \( (\frac{x^{5}}{x^{2}})^{\frac{1}{3}} \)
6. \( (4xy^{2})^{\frac{3}{2}} \)
7. \( \frac{16x^{2}y^{11}z^{5}}{(2xyz)^{6}} \)
8. \( \sqrt[3]{\frac{x^{10}y^{2}}{xy^{14}}} \)

Rationalize the denominator

9. \( \frac{3\sqrt{x}}{2\sqrt{x}-\sqrt{y}} \)
10. \( \frac{1}{\sqrt{x}+2\sqrt{y}} \)

Solutions

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1. Solutions: $$ \begin{align*} & x^{2}y^{3}\cdot x^{3}y= \\[0.5em] & x^{2+3}y^{3+1}= x^{5}y^{4} \end{align*} $$
2. Solutions: $$ \begin{align*} & (xy^{2})^{4}\cdot(x^{2}y)^{3}= \\[0.5em] & (x^{4}y^{8})\cdot(x^{6}y^{3})= x^{10}y^{11} \end{align*} $$
3. \( \frac{(xy)^{5}}{xy^{2}}= x^{5-1}y^{5-2}= x^{4}y^{3} \)
4. Solutions: $$ \begin{align*} & \frac{10xy^{2}z^{3}}{2x^{2}yz}= \\ & 5x^{-1}yz^{2}= \frac{5yz^{2}}{x} \end{align*} $$
5. \( (\frac{x^{5}}{x^{2}})^{\frac{1}{3}}=(x^{3})^{\frac{1}{3}}=x \)
6. Solutions: $$ \begin{align*} & (4xy^{2})^{\frac{3}{2}}= \sqrt{(4xy^{2})^{3}}= \\[0.5em] & \sqrt{64x^{3}y^{6}}=8x^{\frac{3}{2}}y^{3} \end{align*} $$
7. Solutions: $$ \begin{align*} & \frac{16x^{2}y^{11}z^{5}}{(2xyz)^{6}}= \frac{2^{4}x^{2}y^{11}z^{5}}{(2xyz)^{6}}= \\[0.5em] & 2^{-2}x^{-4}y^{5}z^{-1}= \frac{y^{5}}{4x^{4}z} \end{align*} $$
8. \( \sqrt[3]{\frac{x^{10}y^{2}}{xy^{14}}}=\sqrt[3]{\frac{x^{9}}{y^{12}}}=\frac{x^{3}}{y^{4}} \)
9. Solutions: $$ \begin{align*} & \frac{3\sqrt{x}}{2\sqrt{x}-\sqrt{y}}= \\[0.5em] & \frac{3\sqrt{x}}{2\sqrt{x}-\sqrt{y}}\cdot\frac{2\sqrt{x}+\sqrt{y}}{2\sqrt{x}+\sqrt{y}}= \\[0.5em] & \frac{6x+3\sqrt{xy}}{4x-y} \end{align*} $$
10. Solutions: $$ \begin{align*} & \frac{1}{\sqrt{x}+2\sqrt{y}}= \\[0.5em] & \frac{1}{\sqrt{x}+2\sqrt{y}}\cdot\frac{\sqrt{x}-2\sqrt{y}}{\sqrt{x}-2\sqrt{y}}= \\[0.5em] & \frac{\sqrt{x}-2\sqrt{y}}{x-4y} \end{align*} $$

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