 maths > exponents

Arithmetics of Logarithms

what you'll learn...

overview

Arithmetics with logarithms, without evaluating the logarithm, is explained. For example ${\mathrm{log}}_{a}{b}^{m}=m{\mathrm{log}}_{a}b$${\log}_{a} {b}^{m} = m {\log}_{a} b$.

The list of formulas are derived using the first principles of logarithms.

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

We learned that logarithm of a number is the power with which the base is raised to get the number. In other words, logarithm is an inverse of exponent.

${\mathrm{log}}_{2}8={\mathrm{log}}_{2}\left({2}^{3}\right)=3$${\log}_{2} 8 = {\log}_{2} \left({2}^{3}\right) = 3$

By first principles, the value is expressed as an exponent of base
${\mathrm{log}}_{2}8={\mathrm{log}}_{2}{2}^{3}$${\log}_{2} 8 = {\log}_{2} {2}^{3}$

The exponent, to which base is raised to get a number, is the logarithm to the base of the number.
${\mathrm{log}}_{2}8={\mathrm{log}}_{2}{2}^{3}=3$${\log}_{2} 8 = {\log}_{2} {2}^{3} = 3$
$2$$2$ is the base
$8$$8$ is the number
$3$$3$ is the log of the number

The power to which $2$$2$ is raised to get value $8$$8$ is "${\mathrm{log}}_{2}8$${\log}_{2} 8$"

We learned the first principles of logarithm as, for any value $x$$x$ and base $b$$b$,
${\mathrm{log}}_{b}x=n⇔{b}^{n}=x$${\log}_{b} x = n \Leftrightarrow {b}^{n} = x$

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.

log equal implies power equal

It is noted that ${2}^{4}=16$${2}^{4} = 16$ and ${\left(-2\right)}^{4}=16$${\left(- 2\right)}^{4} = 16$, so $4th$$4 t h$ root of $16$$16$ has more than one possible values.
That is generalized as ${p}^{n}={q}^{n}$${p}^{n} = {q}^{n}$ does not imply that $p=q$$p = q$.

If the base $2$$2$ is the same, then to get the value $16$$16$, the exponent is unique $4$$4$. That is only ${2}^{4}=16$${2}^{4} = 16$ for base $2$$2$ and number $16$$16$.

If ${\mathrm{log}}_{b}x={\mathrm{log}}_{b}y$${\log}_{b} x = {\log}_{b} y$ then $x=y$$x = y$.

Note that by first principles,
${\mathrm{log}}_{b}x=n⇔{b}^{n}=x$$\textcolor{c \mathmr{and} a l}{{\log}_{b} x = n \Leftrightarrow {b}^{n} = x}$
${\mathrm{log}}_{b}y=n⇔{b}^{n}=y$$\textcolor{\mathrm{de} e p s k y b l u e}{{\log}_{b} y = n \Leftrightarrow {b}^{n} = y}$

So,
${b}^{n}$$\textcolor{c \mathmr{and} a l}{{b}^{n}}$$={b}^{n}$$= \textcolor{\mathrm{de} e p s k y b l u e}{{b}^{n}}$
$x$$\textcolor{c \mathmr{and} a l}{x}$$=y$$= \textcolor{\mathrm{de} e p s k y b l u e}{y}$

If base of the logarithm is same, then ${\mathrm{log}}_{b}x={\mathrm{log}}_{b}y$${\log}_{b} x = {\log}_{b} y$ implies $x=y$$x = y$.
Generalizing this, for real numbers $x$$x$, $y$$y$ and $b$$b$,
${\mathrm{log}}_{b}x={\mathrm{log}}_{b}y⇔x=y$${\log}_{b} x = {\log}_{b} y \Leftrightarrow x = y$

log of an exponent

${\mathrm{log}}_{2}{2}^{10}$${\log}_{2} {2}^{10}$ $=10$$= 10$
Note that the value is given as an exponent of base.

Generalizing this, for real numbers $b$$b$ and $n$$n$,
${\mathrm{log}}_{b}{b}^{n}=n$${\log}_{b} {b}^{n} = n$

exponent to a log

Consider ${2}^{{\mathrm{log}}_{2}8}=8$${2}^{{\log}_{2} 8} = 8$

Note that by first principles,
${\mathrm{log}}_{2}8=3$$\textcolor{c \mathmr{and} a l}{{\log}_{2} 8 = 3}$
${2}^{3}=8$$\textcolor{\mathrm{de} e p s k y b l u e}{{2}^{3} = 8}$

So,
${2}^{{\mathrm{log}}_{2}8}={2}^{3}$${\textcolor{\mathrm{de} e p s k y b l u e}{2}}^{\textcolor{c \mathmr{and} a l}{{\log}_{2} 8}} = {2}^{3}$

Generalizing this, for real numbers $x$$x$ and $b$$b$,
${b}^{{\mathrm{log}}_{b}x}=x$${b}^{{\log}_{b} x} = x$

log(one)

${\mathrm{log}}_{2}1$${\log}_{2} 1$ $=0$$= 0$
Because, ${2}^{0}=1$${2}^{0} = 1$

Generalizing this, for a real number $b$$b$,
${\mathrm{log}}_{b}1=0$${\log}_{b} 1 = 0$

log(base)

${\mathrm{log}}_{2}2=1$${\log}_{2} 2 = 1$?
Because,${2}^{1}=2$${2}^{1} = 2$

Generalizing this, for a real number $b$$b$,
${\mathrm{log}}_{b}b=1$${\log}_{b} b = 1$

log of a product

It is given that ${\mathrm{log}}_{2}32=5$${\log}_{2} 32 = 5$ and ${\mathrm{log}}_{2}16=4$${\log}_{2} 16 = 4$.

What is ${\mathrm{log}}_{2}\left(32×16\right)$${\log}_{2} \left(32 \times 16\right)$?

By first principles,
${\mathrm{log}}_{2}32=5⇔{2}^{5}=32$$\textcolor{c \mathmr{and} a l}{{\log}_{2} 32 = 5 \Leftrightarrow {2}^{5} = 32}$
${\mathrm{log}}_{2}16=4⇔{2}^{4}=16$$\textcolor{\mathrm{de} e p s k y b l u e}{{\log}_{2} 16 = 4 \Leftrightarrow {2}^{4} = 16}$

So,
${\mathrm{log}}_{2}\left(32$$×16\right)$
$={\mathrm{log}}_{2}\left({2}^{5}$$×{2}^{4}\right)$
substituting the known result ${2}^{5}×{2}^{4}={2}^{5+4}$${2}^{5} \times {2}^{4} = {2}^{5 + 4}$
$={\mathrm{log}}_{2}{2}^{5+4}$$= {\log}_{2} {2}^{\textcolor{c \mathmr{and} a l}{5} + \textcolor{\mathrm{de} e p s k y b l u e}{4}}$
$=5+4$$= \textcolor{c \mathmr{and} a l}{5} + \textcolor{\mathrm{de} e p s k y b l u e}{4}$

That is ${\mathrm{log}}_{2}\left(32$$×16\right)$ $={\mathrm{log}}_{2}\left(32\right)$$= {\log}_{2} \left(\textcolor{c \mathmr{and} a l}{32}\right)$$+{\mathrm{log}}_{2}\left(16\right)$$+ {\log}_{2} \left(\textcolor{\mathrm{de} e p s k y b l u e}{16}\right)$

Generalizing this, for real numbers $x$$x$, $y$$y$ and $b$$b$,
${\mathrm{log}}_{b}\left(xy\right)={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}y$${\log}_{b} \left(x y\right) = {\log}_{b} x + {\log}_{b} y$

log of division

It is given that ${\mathrm{log}}_{2}32=5$${\log}_{2} 32 = 5$ and ${\mathrm{log}}_{2}16=4$${\log}_{2} 16 = 4$.

What is ${\mathrm{log}}_{2}\left(32÷16\right)$${\log}_{2} \left(32 \div 16\right)$?

By first principles,
${\mathrm{log}}_{2}32=5⇔{2}^{5}=32$$\textcolor{c \mathmr{and} a l}{{\log}_{2} 32 = 5 \Leftrightarrow {2}^{5} = 32}$
${\mathrm{log}}_{2}16=4⇔{2}^{4}=16$$\textcolor{\mathrm{de} e p s k y b l u e}{{\log}_{2} 16 = 4 \Leftrightarrow {2}^{4} = 16}$

So,
${\mathrm{log}}_{2}\left(32$$÷16\right)$
${\mathrm{log}}_{2}\left({2}^{5}$$÷{2}^{4}\right)$
substituting the known result ${2}^{5}÷{2}^{4}={2}^{5-4}$${2}^{5} \div {2}^{4} = {2}^{5 - 4}$
${\mathrm{log}}_{2}{2}^{5-4}$${\log}_{2} {2}^{\textcolor{c \mathmr{and} a l}{5} - \textcolor{\mathrm{de} e p s k y b l u e}{4}}$
$=5-4$$= \textcolor{c \mathmr{and} a l}{5} - \textcolor{\mathrm{de} e p s k y b l u e}{4}$".

That is ${\mathrm{log}}_{2}\left(32$$÷16\right)$ $={\mathrm{log}}_{2}\left(32\right)$$= {\log}_{2} \left(\textcolor{c \mathmr{and} a l}{32}\right)$$-{\mathrm{log}}_{2}\left(16\right)$$- {\log}_{2} \left(\textcolor{\mathrm{de} e p s k y b l u e}{16}\right)$

Generalizing this, for real numbers $x$$x$, $y$$y$ and $b$$b$,
${\mathrm{log}}_{b}\left(\frac{x}{y}\right)={\mathrm{log}}_{b}x-{\mathrm{log}}_{b}y$

log(exponent) again

What is ${\mathrm{log}}_{2}\left({8}^{5}\right)$${\log}_{2} \left({8}^{5}\right)$?

By first principles,
${\mathrm{log}}_{2}8=3⇔{2}^{3}=8$$\textcolor{c \mathmr{and} a l}{{\log}_{2} 8 = 3 \Leftrightarrow {2}^{3} = 8}$

${\mathrm{log}}_{2}{8}^{5}$${\log}_{2} {\textcolor{c \mathmr{and} a l}{8}}^{5}$
$={\mathrm{log}}_{2}{\left({2}^{3}\right)}^{5}\right)$
$={\mathrm{log}}_{2}{2}^{3×5}$$= {\log}_{2} {2}^{\textcolor{c \mathmr{and} a l}{3} \times 5}$
$=3×5$$= \textcolor{c \mathmr{and} a l}{3} \times 5$
$=5×3$$= 5 \times \textcolor{c \mathmr{and} a l}{3}$
substituting that ${\mathrm{log}}_{2}8=3$${\log}_{2} 8 = 3$
$=5×{\mathrm{log}}_{2}8$$= 5 \times \textcolor{c \mathmr{and} a l}{{\log}_{2} 8}$

Generalizing this, for real numbers $x$$x$,$n$$n$ and $b$$b$,
${\mathrm{log}}_{b}{x}^{n}=n{\mathrm{log}}_{b}x$${\log}_{b} {x}^{n} = n {\log}_{b} x$

It is noted that
${b}^{n}=x⇔{\mathrm{log}}_{b}x=n$${b}^{n} = x \Leftrightarrow {\log}_{b} x = n$

${2}^{3}=8⇔\sqrt{8}={8}^{\frac{1}{3}}=2$$\textcolor{c \mathmr{and} a l}{{2}^{3} = 8} \Leftrightarrow \sqrt{8} = \textcolor{\mathrm{de} e p s k y b l u e}{{8}^{\frac{1}{3}} = 2}$
There are two exponents in the above ${2}^{3}$$\textcolor{c \mathmr{and} a l}{{2}^{3}}$ and ${8}^{\frac{1}{3}}$$\textcolor{\mathrm{de} e p s k y b l u e}{{8}^{\frac{1}{3}}}$

applying logarithm for the exponent ${2}^{3}=8$$\textcolor{c \mathmr{and} a l}{{2}^{3} = 8}$ with $b=2$$b = 2$ and $x=8$$x = 8$, we get
${\mathrm{log}}_{2}8=3$$\textcolor{c \mathmr{and} a l}{{\log}_{2} 8 = 3}$

applying logarithm for the exponent ${8}^{\frac{1}{3}}=2$$\textcolor{\mathrm{de} e p s k y b l u e}{{8}^{\frac{1}{3}} = 2}$ with $b=8$$b = 8$ and $x=2$$x = 2$
${\mathrm{log}}_{8}2=\frac{1}{3}$$\textcolor{\mathrm{de} e p s k y b l u e}{{\log}_{8} 2 = \frac{1}{3}}$

Comparing the two logarithms, "${\mathrm{log}}_{2}8=1/{\mathrm{log}}_{8}2$${\log}_{2} 8 = 1 / {\log}_{8} 2$".

Generalizing this, for real numbers $a$$a$ and $b$$b$,
${\mathrm{log}}_{b}a=1÷\left({\mathrm{log}}_{a}b\right)$${\log}_{b} a = 1 \div \left({\log}_{a} b\right)$

log of different base

It is noted that

${2}^{12}={4}^{6}$$\textcolor{c \mathmr{and} a l}{{2}^{12}} = \textcolor{\mathrm{de} e p s k y b l u e}{{4}^{6}}$

Converting the left hand side ${2}^{12}$$\textcolor{c \mathmr{and} a l}{{2}^{12}}$ to log form
${\mathrm{log}}_{2}\left({2}^{12}\right)=12$$\textcolor{c \mathmr{and} a l}{{\log}_{2} \left({2}^{12}\right) = 12}$

The same in right hand side
${\mathrm{log}}_{2}\left({4}^{6}\right)=6×{\mathrm{log}}_{2}\left(4\right)$$\textcolor{c \mathmr{and} a l}{{\log}_{2} \left({4}^{6}\right)} = \textcolor{\mathrm{de} e p s k y b l u e}{6} \times \textcolor{c \mathmr{and} a l}{{\log}_{2} \left(4\right)}$

Converting the right hand side ${4}^{6}$$\textcolor{\mathrm{de} e p s k y b l u e}{{4}^{6}}$ to log form
${\mathrm{log}}_{4}\left({4}^{6}\right)=6$$\textcolor{\mathrm{de} e p s k y b l u e}{{\log}_{4} \left({4}^{6}\right) = 6}$

substituting the value $6$$6$
${\mathrm{log}}_{2}\left({4}^{6}\right)={\mathrm{log}}_{4}\left({4}^{6}\right)×{\mathrm{log}}_{2}\left(4\right)$$\textcolor{c \mathmr{and} a l}{{\log}_{2} \left({4}^{6}\right)} = \textcolor{\mathrm{de} e p s k y b l u e}{{\log}_{4} \left({4}^{6}\right)} \times \textcolor{c \mathmr{and} a l}{{\log}_{2} \left(4\right)}$

The above proves
${\mathrm{log}}_{2}\left({2}^{12}\right)={\mathrm{log}}_{4}\left({2}^{12}\right)×{\mathrm{log}}_{2}4$${\log}_{2} \left({2}^{12}\right) = {\log}_{4} \left({2}^{12}\right) \times {\log}_{2} 4$
${\mathrm{log}}_{2}\left({2}^{12}\right)={\mathrm{log}}_{4}\left({2}^{12}\right)÷{\mathrm{log}}_{4}2$${\log}_{2} \left({2}^{12}\right) = {\log}_{4} \left({2}^{12}\right) \div {\log}_{4} 2$

Generalizing this, for real numbers $x$$x$, $a$$a$ and $b$$b$,
${\mathrm{log}}_{b}x={\mathrm{log}}_{a}x×{\mathrm{log}}_{b}a$${\log}_{b} x = {\log}_{a} x \times {\log}_{b} a$
${\mathrm{log}}_{b}x={\mathrm{log}}_{a}x÷{\mathrm{log}}_{a}b$${\log}_{b} x = {\log}_{a} x \div {\log}_{a} b$

summary

Known results in Logarithms :

${\mathrm{log}}_{b}x=n⇔{b}^{n}=x$${\log}_{b} x = n \Leftrightarrow {b}^{n} = x$

${\mathrm{log}}_{b}x={\mathrm{log}}_{b}y⇔x=y$${\log}_{b} x = {\log}_{b} y \Leftrightarrow x = y$

${\mathrm{log}}_{b}{b}^{n}=n$${\log}_{b} {b}^{n} = n$

${b}^{{\mathrm{log}}_{b}x}=x$${b}^{{\log}_{b} x} = x$

${\mathrm{log}}_{b}1=0$${\log}_{b} 1 = 0$

${\mathrm{log}}_{b}b=1$${\log}_{b} b = 1$

${\mathrm{log}}_{b}\left(xy\right)={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}y$${\log}_{b} \left(x y\right) = {\log}_{b} x + {\log}_{b} y$

${\mathrm{log}}_{b}\left(\frac{x}{y}\right)={\mathrm{log}}_{b}x-{\mathrm{log}}_{b}y$

${\mathrm{log}}_{b}{x}^{n}=n{\mathrm{log}}_{b}x$${\log}_{b} {x}^{n} = n {\log}_{b} x$

${\mathrm{log}}_{b}x=\left({\mathrm{log}}_{a}x\right)÷\left({\mathrm{log}}_{a}b\right)$${\log}_{b} x = \left({\log}_{a} x\right) \div \left({\log}_{a} b\right)$

${\mathrm{log}}_{b}x={\mathrm{log}}_{b}a×{\mathrm{log}}_{a}x$${\log}_{b} x = {\log}_{b} a \times {\log}_{a} x$

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