Laws of Indices (DP IB Analysis & Approaches (AA): SL): Revision Note

Law

Description

How it works

$a^1=a$

Anything to the power of 1 is itself

$6^1=6$

$a^0=1$

Anything to the power of 0 is 1

$8^0=1$

$a^m\times a^n=a^{m+n}$

To multiply indices with the same base, add their powers

$$\begin{gathered} 4^3\times 4^2 \\ =\left(4\times 4\times 4\right)\times \left(4\times 4\right) \\ =4^5 \end{gathered}$$

$a^m÷a^n=\frac{a^m}{a^n}=a^{m-n}$

To divide indices with the same base, subtract their powers

$$\begin{gathered} 7^5÷7^2 \\ =\frac{7\times 7\times 7\times \cancel{7}\times \cancel{7}}{\cancel{7}\times \cancel{7}} \\ =7^3 \end{gathered}$$

${\left(a^m\right)}^n=a^{mn}$

To raise indices to a new power, multiply their powers

$$\begin{gathered} {\left({14}^3\right)}^2 \\ =\left(14\times 14\times 14\right)\times \left(14\times 14\times 14\right) \\ ={14}^6 \end{gathered}$$

${\left(ab\right)}^n=a^n{b}^n$

To raise a product to a power, apply the power to both numbers, and multiply

${\left(3\times 4\right)}^2=3^2\times 4^2$

${\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}$

To raise a fraction to a power, apply the power to both the numerator and denominator

${\left(\frac{3}{4}\right)}^2=\frac{3^2}{4^2}=\frac{9}{16}$

$a^{-1}=\frac{1}{a}$

$a^{-n}=\frac{1}{a^n}$

A negative power is the reciprocal

$6^{-1}=\frac{1}{6}$

${11}^{-3}=\frac{1}{{11}^3}$

${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n=\frac{b^n}{a^n}$

A fraction to a negative power, is the reciprocal of the fraction, to the positive power

${\left(\frac{2}{5}\right)}^{-3}={\left(\frac{5}{2}\right)}^3=\frac{5^3}{2^3}=\frac{125}{8}$

$a^{\frac{1}{n}}=\sqrt[n]{a}$

The fractional power $\frac{1}{n}$ is the n^th root ( )

${25}^{\frac{1}{2}}=\sqrt[2]{25}=5$

${27}^{\frac{1}{3}}=\sqrt[3]{27}=3$

$$\begin{gathered} a^{-\frac{1}{n}}={\left(a^{\frac{1}{n}}\right)}^{-1} \\ ={\left(\sqrt[n]{a}\right)}^{-1}=\frac{1}{\sqrt[n]{a}} \end{gathered}$$

A negative, fractional power is one over a root

${64}^{-\frac{1}{2}}=\frac{1}{\sqrt[2]{64}}=\frac{1}{8}$

${125}^{-\frac{1}{3}}=\frac{1}{\sqrt[3]{125}}=\frac{1}{5}$

$$\begin{gathered} a^{\frac{m}{n}}=a^{\frac{1}{n}\times m} \\ ={\left(a^{\frac{1}{n}}\right)}^m={\left(a^m\right)}^{\frac{1}{n}} \end{gathered}$$

The fractional power $\frac{m}{n}$ is the n^th root all to the power m, , or the n^th root of the power m, (both are the same)

$8^{\frac{2}{3}}={\left(8^{\frac{1}{3}}\right)}^2={\left(\sqrt[3]{8}\right)}^2=2^2=4$

$8^{\frac{2}{3}}={\left(8^2\right)}^{\frac{1}{3}}=\sqrt[3]{64}=4$