 maths > exponents

Arithmetics with Exponents

what you'll learn...

overview

Arithmetics with exponents, without evaluating the exponent, is explained. For example ${a}^{m}×{a}^{n}={a}^{m+n}$${a}^{m} \times {a}^{n} = {a}^{m + n}$. The list of formulas are derived using the first principles of exponent.

These known results are given as a set of formulas. Students are advised to work them out quickly using the first principles. No need to memorize, and if the formulas are used repeatedly, over time, these can be recalled quickly.

first principles for all formulae

"base" repeatedly multiplied the "power" number of times equals "exponent".

The value of ${2}^{5}$${2}^{5}$ is "${2}^{5}=2×2×2×2×2$${2}^{5} = 2 \times 2 \times 2 \times 2 \times 2$".

We learned the first principles of exponents as, for any numbers $a$$a$ and $n$$n$,
${a}^{n}=a×a×a×\cdots \left(n\phantom{\rule{1ex}{0ex}}\text{times}\right)$${a}^{n} = a \times a \times a \times \cdots \left(n \textrm{\times}\right)$

In this topic, the above is used to understand some known results.

These known results are given as a set of formulas. Students are advised to work them out quickly using the first principles. No need to memorize, and if the formulas are used repeatedly, over time, these can be recalled quickly.

multiplication of exponents of same base

Consider ${2}^{4}×{2}^{3}$${2}^{4} \times {2}^{3}$?

Note that, by first principles
${2}^{4}=2×2×2×2$$\textcolor{c \mathmr{and} a l}{{2}^{4} = 2 \times 2 \times 2 \times 2}$
${2}^{3}=2×2×2$$\textcolor{\mathrm{de} e p s k y b l u e}{{2}^{3} = 2 \times 2 \times 2}$

So, the given expression
${2}^{4}×{2}^{3}$$\textcolor{c \mathmr{and} a l}{{2}^{4}} \times \textcolor{\mathrm{de} e p s k y b l u e}{{2}^{3}}$
$=2×2×2×2×2×2×2$$= \textcolor{c \mathmr{and} a l}{2 \times 2 \times 2 \times 2} \times \textcolor{\mathrm{de} e p s k y b l u e}{2 \times 2 \times 2}$
the simplified expression is ${2}^{4+3}={2}^{7}$${2}^{4 + 3} = {2}^{7}$

Generalizing this, for real numbers $a$$a$, $m$$m$ and $n$$n$,
${a}^{m}×{a}^{n}={a}^{m+n}$${a}^{m} \times {a}^{n} = {a}^{m + n}$

multiplication of exponents of diff base

Consider ${2}^{4}×{3}^{4}$${2}^{4} \times {3}^{4}$?

Note : by first principles
${2}^{4}=2×2×2×2$$\textcolor{c \mathmr{and} a l}{{2}^{4} = 2 \times 2 \times 2 \times 2}$
${3}^{4}=3×3×3×3$$\textcolor{\mathrm{de} e p s k y b l u e}{{3}^{4} = 3 \times 3 \times 3 \times 3}$

So, the given expression
${2}^{4}×{3}^{4}$$\textcolor{c \mathmr{and} a l}{{2}^{4}} \times \textcolor{\mathrm{de} e p s k y b l u e}{{3}^{4}}$
$=2×2×2×2×3×3×3×3$$= \textcolor{c \mathmr{and} a l}{2 \times 2 \times 2 \times 2} \times \textcolor{\mathrm{de} e p s k y b l u e}{3 \times 3 \times 3 \times 3}$
$=\left(2×3\right)$$= \left(\textcolor{c \mathmr{and} a l}{2} \times \textcolor{\mathrm{de} e p s k y b l u e}{3}\right)$$×\left(2×3\right)$$\times \left(\textcolor{c \mathmr{and} a l}{2} \times \textcolor{\mathrm{de} e p s k y b l u e}{3}\right)$$×\left(2×3\right)$$\times \left(\textcolor{c \mathmr{and} a l}{2} \times \textcolor{\mathrm{de} e p s k y b l u e}{3}\right)$$×\left(2×3\right)$$\times \left(\textcolor{c \mathmr{and} a l}{2} \times \textcolor{\mathrm{de} e p s k y b l u e}{3}\right)$
The simplified form of the expression is "${\left(2×3\right)}^{4}$${\left(2 \times 3\right)}^{4}$"

Generalizing this, for real numbers $a$$a$, $m$$m$ and $n$$n$,
${a}^{m}×{b}^{m}={\left(a×b\right)}^{m}$${a}^{m} \times {b}^{m} = {\left(a \times b\right)}^{m}$

division of exponents of same base

Consider ${2}^{4}÷{2}^{3}$${2}^{4} \div {2}^{3}$

Note : by first principles
${2}^{4}=2×2×2×2$$\textcolor{c \mathmr{and} a l}{{2}^{4} = 2 \times 2 \times 2 \times 2}$
${2}^{3}=2×2×2$$\textcolor{\mathrm{de} e p s k y b l u e}{{2}^{3} = 2 \times 2 \times 2}$

So, the given expression
${2}^{4}÷{2}^{3}$$\textcolor{c \mathmr{and} a l}{{2}^{4}} \div \textcolor{\mathrm{de} e p s k y b l u e}{{2}^{3}}$
$=2×2×2×2÷\left[2×2×2\right]$$= \textcolor{c \mathmr{and} a l}{2 \times 2 \times 2 \times 2} \div \left[\textcolor{\mathrm{de} e p s k y b l u e}{2 \times 2 \times 2}\right]$
by properties of arithmetics
$=2×2×2×2×\left[\frac{1}{2}×\frac{1}{2}×\frac{1}{2}\right]$$= \textcolor{c \mathmr{and} a l}{2 \times 2 \times 2 \times 2} \times \left[\textcolor{\mathrm{de} e p s k y b l u e}{\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}}\right]$
The simplified expression is ${2}^{4-3}={2}^{1}$${2}^{4 - 3} = {2}^{1}$

Generalizing this, for real numbers $a$$a$, $m$$m$ and $n$$n$,
${a}^{m}÷{a}^{n}={a}^{m-n}$${a}^{m} \div {a}^{n} = {a}^{m - n}$

division of exponents of diff base

Consider ${2}^{4}÷{3}^{4}$${2}^{4} \div {3}^{4}$

Note : by first principles
${2}^{4}=2×2×2×2$$\textcolor{c \mathmr{and} a l}{{2}^{4} = 2 \times 2 \times 2 \times 2}$
${3}^{4}=3×3×3×3$$\textcolor{\mathrm{de} e p s k y b l u e}{{3}^{4} = 3 \times 3 \times 3 \times 3}$

So, the given expression
${2}^{4}÷{3}^{4}$$\textcolor{c \mathmr{and} a l}{{2}^{4}} \div \textcolor{\mathrm{de} e p s k y b l u e}{{3}^{4}}$
$=2×2×2×2÷3×3×3×3$$= \textcolor{c \mathmr{and} a l}{2 \times 2 \times 2 \times 2} \div \textcolor{\mathrm{de} e p s k y b l u e}{3 \times 3 \times 3 \times 3}$
$=2×2×2×2×\frac{1}{3}×\frac{1}{3}×\frac{1}{3}×\frac{1}{3}$$= \textcolor{c \mathmr{and} a l}{2 \times 2 \times 2 \times 2} \times \textcolor{\mathrm{de} e p s k y b l u e}{\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}}$
$=\frac{2}{3}×\frac{2}{3}$$= \frac{\textcolor{c \mathmr{and} a l}{2}}{\textcolor{\mathrm{de} e p s k y b l u e}{3}} \times \frac{\textcolor{c \mathmr{and} a l}{2}}{\textcolor{\mathrm{de} e p s k y b l u e}{3}}$$×\frac{2}{3}×\frac{2}{3}$$\times \frac{\textcolor{c \mathmr{and} a l}{2}}{\textcolor{\mathrm{de} e p s k y b l u e}{3}} \times \frac{\textcolor{c \mathmr{and} a l}{2}}{\textcolor{\mathrm{de} e p s k y b l u e}{3}}$
The simplified expression is ${\left(\frac{2}{3}\right)}^{4}$${\left(\frac{2}{3}\right)}^{4}$

Generalizing this, for real numbers $a$$a$, $m$$m$ and $n$$n$,
${a}^{m}÷{b}^{m}={\left(a÷b\right)}^{m}$${a}^{m} \div {b}^{m} = {\left(a \div b\right)}^{m}$

exponents of power 0

Consider ${3}^{0}$${3}^{0}$

Note that
${3}^{0}$${3}^{0}$
$={3}^{1-1}$$= {3}^{1 - 1}$
$={3}^{1}÷{3}^{1}$$= {3}^{1} \div {3}^{1}$
$=\frac{3}{3}$$= \frac{3}{3}$
The value of the expression is "1".

Generalizing this, for a real number $a$$a$,
${a}^{0}=1$${a}^{0} = 1$

exponents of exponent

Consider ${\left({5}^{2}\right)}^{3}$${\left({5}^{2}\right)}^{3}$

Note that by first principles,
${\left({5}^{2}\right)}^{3}$${\left({5}^{2}\right)}^{3}$
$=\left({5}^{2}\right)×\left({5}^{2}\right)×\left({5}^{2}\right)$$= \left({5}^{2}\right) \times \left({5}^{2}\right) \times \left({5}^{2}\right)$
substituting ${5}^{2}=5×5$${5}^{2} = 5 \times 5$
$=\left(5×5\right)×\left(5×5\right)×\left(5×5\right)$$= \left(5 \times 5\right) \times \left(5 \times 5\right) \times \left(5 \times 5\right)$

The simplified expression is ${5}^{2×3}={5}^{6}$${5}^{2 \times 3} = {5}^{6}$

Generalizing this, for real numbers $a$$a$, $m$$m$ and $n$$n$,
${\left({a}^{m}\right)}^{n}={a}^{mn}$${\left({a}^{m}\right)}^{n} = {a}^{m n}$

consider ${2}^{3}+{2}^{3}+{2}^{3}+{2}^{3}+{2}^{3}$${2}^{3} + {2}^{3} + {2}^{3} + {2}^{3} + {2}^{3}$

Note, by first principles a number repeatedly added is a multiplication, and so it is "$5×{2}^{3}$$5 \times {2}^{3}$"

Generalizing this, for real numbers $a$$a$, $m$$m$ and $n$$n$,
${a}^{m}+{a}^{m}+\cdots \left(n\text{times}\right)=n×{a}^{m}$${a}^{m} + {a}^{m} + \cdots \left(n \textrm{\times}\right) = n \times {a}^{m}$

Consider $3×{2}^{5}+4×{2}^{5}$$3 \times {2}^{5} + 4 \times {2}^{5}$

Note that by first principles,
$3×{2}^{5}={2}^{5}+{2}^{5}+{2}^{5}$$\textcolor{c \mathmr{and} a l}{3 \times {2}^{5} = {2}^{5} + {2}^{5} + {2}^{5}}$
$4×{2}^{5}={2}^{5}+{2}^{5}+{2}^{5}+{2}^{5}$$\textcolor{\mathrm{de} e p s k y b l u e}{4 \times {2}^{5} = {2}^{5} + {2}^{5} + {2}^{5} + {2}^{5}}$

So,
$3×{2}^{5}$$\textcolor{c \mathmr{and} a l}{3 \times {2}^{5}}$$+4×{2}^{5}$$+ \textcolor{\mathrm{de} e p s k y b l u e}{4 \times {2}^{5}}$
$={2}^{5}+{2}^{5}+{2}^{5}$$= \textcolor{c \mathmr{and} a l}{{2}^{5} + {2}^{5} + {2}^{5}}$$+{2}^{5}+{2}^{5}+{2}^{5}+{2}^{5}$$+ \textcolor{\mathrm{de} e p s k y b l u e}{{2}^{5} + {2}^{5} + {2}^{5} + {2}^{5}}$

The simplified form of the above expression is "$\left(3+4\right)×{2}^{5}=7×{2}^{5}$$\left(3 + 4\right) \times {2}^{5} = 7 \times {2}^{5}$".

Generalizing this, for real numbers $a$$a$, $m$$m$, $p$$p$ and $q$$q$,
$p×{a}^{m}+q×{a}^{m}=\left(p+q\right)×{a}^{m}$$p \times {a}^{m} + q \times {a}^{m} = \left(p + q\right) \times {a}^{m}$

summary

Known results in Exponents :
For $a,b,p,q,m,n\in ℝ$$a , b , p , q , m , n \in \mathbb{R}$,

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

${a}^{m}×{b}^{m}={\left(a×b\right)}^{m}$${a}^{m} \times {b}^{m} = {\left(a \times b\right)}^{m}$

${a}^{m}÷{a}^{n}={a}^{m-n}$${a}^{m} \div {a}^{n} = {a}^{m - n}$

${a}^{m}÷{b}^{m}={\left(a÷b\right)}^{m}$${a}^{m} \div {b}^{m} = {\left(a \div b\right)}^{m}$

${a}^{0}=1$${a}^{0} = 1$

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

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

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

${a}^{m}+{a}^{m}+\cdots \left(n×\right)=n×{a}^{m}$${a}^{m} + {a}^{m} + \cdots \left(n \times\right) = n \times {a}^{m}$

$p×{a}^{m}+q×{a}^{m}=\left(p+q\right)×{a}^{m}$$p \times {a}^{m} + q \times {a}^{m} = \left(p + q\right) \times {a}^{m}$

Outline

The outline of material to learn "Exponents" is as follows. Note: click here for detailed outline of Exponents s

→   Representation of Exponents

→   Inverse of exponent : root

→   Inverse of exponent : Logarithm

→   Common and Natural Logarithms

→   Exponents Arithmetics

→   Logarithm Arithmetics

→   Formulas

→   Numerical Expressions

→   PEMA / BOMA

→   Squares and Square roots

→   Cubes and Cube roots