The laws of powers and a rational exponent

Exponential and logarithmic functions, grade 12 (5 units) Lesson 1 of 23Learning

The laws of powers and a rational exponent

The laws of powers

In a power $\displaystyle a^x$ , $\displaystyle a$ is the base and $\displaystyle x$ the exponent. The laws of powers turn multiplying and dividing powers with the same base into adding and subtracting their exponents.

\displaystyle a^{x}\cdot a^{y}

\displaystyle =

\displaystyle a^{x+y}

\displaystyle \frac{a^{x}}{a^{y}}

\displaystyle =

\displaystyle a^{x-y}

\displaystyle (a^{x})^{y}

\displaystyle =

\displaystyle a^{x\cdot y}

\displaystyle (a\cdot b)^{x}

\displaystyle =

\displaystyle a^{x}\cdot b^{x}

\displaystyle \left(\frac{a}{b}\right)^{x}

\displaystyle =

\displaystyle \frac{a^{x}}{b^{x}}

The exponents 0 and negative

The quotient law decides what $\displaystyle a^0$ and $\displaystyle a^{-x}$ are: $\displaystyle {\frac{a^{5}}{a^{5}}}={a^{0}}$ is 1, and $\displaystyle {\frac{a^{2}}{a^{6}}}={a^{-4}}$ is $\displaystyle \frac{1}{a^{4}}$ .

\displaystyle a^{0}

\displaystyle =

\displaystyle 1

\displaystyle a\neq0

\displaystyle a^{-x}

\displaystyle =

\displaystyle \frac{1}{a^{x}}

$\displaystyle {5^{-2}}={\frac{1}{25}}$ .
$\displaystyle {\left(\frac{3}{4}\right)^{-1}}={\frac{4}{3}}$ : a negative exponent turns the fraction over.

A rational exponent

A fraction in the exponent is a root: the fraction’s denominator is the root’s index, its numerator the power.

\displaystyle a^{\frac{m}{n}}

\displaystyle =

\displaystyle \sqrt[n]{a^{m}}

\displaystyle a>0

$\displaystyle {32^{\frac{2}{5}}}={\left(\sqrt[5]{32}\right)^{2}}={2^{2}}={4}$ : it is easier to take the root first, and only then the power.
$\displaystyle {9^{-\frac{3}{2}}}={\frac{1}{\left(\sqrt{9}\right)^{3}}}={\frac{1}{27}}$ .
Every root is a power: $\displaystyle {\sqrt{x}}={x^{\frac{1}{2}}}$ , $\displaystyle {\sqrt[3]{x^{2}}}={x^{\frac{2}{3}}}$ , $\displaystyle {\frac{1}{\sqrt[4]{x}}}={x^{-\frac{1}{4}}}$ .
All the laws of powers hold for rational exponents too.
On the formula sheet the base is positive. When the fraction is in lowest terms and its denominator is odd, a negative base has a value too: $\displaystyle {(-8)^{\frac{1}{3}}}={\sqrt[3]{-8}}={-2}$ .

A common base

Powers with different bases become powers of one base: $\displaystyle {4}={2^2}$ , $\displaystyle {8}={2^3}$ , $\displaystyle {\frac{1}{16}}={2^{-4}}$ , $\displaystyle {\sqrt{2}}={2^{\frac{1}{2}}}$ . Then the laws of powers work.

$\displaystyle {8^{x}}={(2^{3})^{x}}={2^{3x}}$ , and $\displaystyle {\frac{1}{4^{x}}}={2^{-2x}}$ .
Comparing: which is greater, $\displaystyle 2^{30}$ or $\displaystyle 3^{20}$ ? Write both with the exponent 10: $\displaystyle {2^{30}}={(2^{3})^{10}}={8^{10}}$ and $\displaystyle {3^{20}}={(3^{2})^{10}}={9^{10}}$ . So $\displaystyle 3^{20}>2^{30}$ .

Worked example: a number

The example below computes $\displaystyle \frac{4^{\frac{3}{2}}\cdot8^{-\frac{1}{3}}}{16^{\frac{1}{4}}}$ , writing every base as a power of 2:

\frac{4^{\frac{3}{2}}\cdot8^{-\frac{1}{3}}}{16^{\frac{1}{4}}}=\frac{(2^{2})^{\frac{3}{2}}\cdot(2^{3})^{-\frac{1}{3}}}{(2^{4})^{\frac{1}{4}}}=\frac{2^{3}\cdot2^{-1}}{2^{1}}=2^{3-1-1}=2

Worked example: letters

The example below simplifies an expression in which $\displaystyle a>0$ . It writes the roots as powers, then adds and subtracts the exponents:

\frac{\sqrt[3]{a^{2}}\cdot\sqrt{a}}{a^{\frac{1}{6}}}=\frac{a^{\frac{2}{3}}\cdot a^{\frac{1}{2}}}{a^{\frac{1}{6}}}=a^{\frac{2}{3}+\frac{1}{2}-\frac{1}{6}}=a^{\frac{4+3-1}{6}}=a^{1}=a

Exercise 1

Calculate.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power