 maths > exponents

Understanding exponents, roots, logarithms

what you'll learn...

overview

This topic is little advanced for high school students. The various possibilities for roots and logarithms are discussed.

•  root with negative power ($\sqrt[-2]{4}$$\sqrt[- 2]{4}$)

•  root with power 0 ($\sqrt{4}$$\sqrt{4}$)

•  logarithm with fraction base (${\mathrm{log}}_{\frac{1}{2}}8$${\log}_{\frac{1}{2}} 8$)

•  logarithm of negative value (${\mathrm{log}}_{2}\left(-4\right)$${\log}_{2} \left(- 4\right)$)

•  logarithm with negative base (${\mathrm{log}}_{-2}\left(-8\right)$${\log}_{- 2} \left(- 8\right)$)

•  logarithm with 0 base (${\mathrm{log}}_{0}2$${\log}_{0} 2$)

•  logarithm with 1 base (${\mathrm{log}}_{1}2$${\log}_{1} 2$)

Note: Some of the listed ones are not defined and has no meaning. Read through the lesson to understand the details.

first principles

revising the basics

Exponent is
${\left(\text{base}\right)}^{\text{power}}=\text{exp. result}$${\left(\textrm{b a s e}\right)}^{\textrm{p o w e r}} = \textrm{\exp . r e s \underline{t}}$

If exponentiation result and power are given, then
$\text{base}=\sqrt[\text{power}]{\text{result}}$$\textrm{b a s e} = \sqrt[\textrm{p o w e r}]{\textrm{r e s \underline{t}}}$
This inverse is called the "root".

If exponentiation-result and base are given, then
$\text{power}={\mathrm{log}}_{\text{base}}\left(\text{exp.result}\right)$$\textrm{p o w e r} = {\log}_{\textrm{b a s e}} \left(\textrm{\exp . r e s \underline{t}}\right)$
This inverse is called the "logarithm".

all combinations and their uses

Consider ${\left(\text{base}\right)}^{\text{power}}=\text{exp. result}$${\left(\textrm{b a s e}\right)}^{\textrm{p o w e r}} = \textrm{\exp . r e s \underline{t}}$

$\text{base}$$\textrm{b a s e}$ and $\text{power}$$\textrm{p o w e r}$ can be one of positive integers, negative integers, $0$$0$, $1$$1$, or fractions.

Let us consider the important combinations one by one and understand some properties of the inverses "root" and "logarithm".

positive is simple enough

Consider ${\left(\text{base}\right)}^{\text{power}}=\text{exp. result}$${\left(\textrm{b a s e}\right)}^{\textrm{p o w e r}} = \textrm{\exp . r e s \underline{t}}$

Given that $\text{base}$$\textrm{b a s e}$ and $\text{power}$$\textrm{p o w e r}$ are positive integers. The $\text{exp. result}$$\textrm{\exp . r e s \underline{t}}$ a positive integer".

negative base - not so simple

Given that, $\text{base}$$\textrm{b a s e}$ is a negative integer and $\text{power}$$\textrm{p o w e r}$ is a positive integer. The $\text{exp. result}$$\textrm{\exp . r e s \underline{t}}$ can be a positive or negative value".

The negative value of the base is shown as $-a$$- a$ with a positive integer $a$$a$.

The exponentiation is
${\left(-a\right)}^{b}=d$${\left(- a\right)}^{b} = d$
The result $d$$d$ can be a positive or a negative value.

The equation can be simplified to
${\left(-a\right)}^{b}={\left(-1\right)}^{b}{a}^{b}={\left(-1\right)}^{b}c$${\left(- a\right)}^{b} = {\left(- 1\right)}^{b} {a}^{b} = {\left(- 1\right)}^{b} c$
and so, the value $c$$c$ is always positive.

In dealing with a negative base, the sign is separated out.

The following are noted.

•  In the inverse "root", the exponentiation result (number $c$$c$) is a positive number. Negative values for exponentiation result are studied separately in the higher level mathematics (complex numbers).

•  In the inverse "logarithm", the base (number $a$$a$) is a positive number. Negative bases are not studied, as the negative base is handled as power of $-1$$- 1$ as shown.

•  In the inverse "logarithm", the exponentiation result (number $c$$c$) is a positive number. $\mathrm{log}$$\log$ of negative values are not defined.

Note 1: If the $\text{base}$$\textrm{b a s e}$ is a positive value, then the $\text{exp. result}$$\textrm{\exp . r e s \underline{t}}$ is always positive. So logarithm of negative values is not defined.

Note 2: If we encounter ${\mathrm{log}}_{\text{-ve}\phantom{\rule{1ex}{0ex}}a}\left(\text{+ve}\phantom{\rule{1ex}{0ex}}c\right)=b$${\log}_{\textrm{- v e} a} \left(\textrm{+ v e} c\right) = b$ or ${\mathrm{log}}_{\text{-ve}\phantom{\rule{1ex}{0ex}}a}\left(\text{-ve}\phantom{\rule{1ex}{0ex}}c\right)=b$${\log}_{\textrm{- v e} a} \left(\textrm{- v e} c\right) = b$, then convert them into the equivalent exponent equation ${a}^{b}=c$${a}^{b} = c$ and solve for $b$$b$.

negative power

Given that $\text{base}$$\textrm{b a s e}$ is a positive integer and $\text{power}$$\textrm{p o w e r}$ is a negative integer. $\text{exp. result}$$\textrm{\exp . r e s \underline{t}}$ is "a positive fraction".

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

So, the result is a positive fraction.

Given that $\text{base}$$\textrm{b a s e}$ is a positive integer and $\text{power}$$\textrm{p o w e r}$ is a negative integer.

The negative value of power is shown as $-b$$- b$ with a positive integer $b$$b$.

The exponentiation is
${a}^{-b}=\frac{1}{{a}^{b}}={\left(\frac{1}{a}\right)}^{b}=c=\frac{1}{d}$${a}^{- b} = \frac{1}{{a}^{b}} = {\left(\frac{1}{a}\right)}^{b} = c = \frac{1}{d}$
The result $c$$c$ is a positive fraction.

The following are noted.

•  In the inverse "root", the power is always a positive number. Roots with negative powers are not studied.

Note 1: If root to the negative power is encountered, then convert that into its equivalent exponentiation equation ${a}^{b}=c$${a}^{b} = c$ and solve for the unknown value.

Note 2: In the inverse "logarithm", the base can be a fraction as given in the equation with $\frac{1}{a}$$\frac{1}{a}$. This is explained along with the generic form of base $\frac{p}{q}$$\frac{p}{q}$ in the next page.

fraction as base

Given that $\text{base}$$\textrm{b a s e}$ is a positive fraction.
${a}^{b}={\left(\frac{p}{q}\right)}^{b}=c$${a}^{b} = {\left(\frac{p}{q}\right)}^{b} = c$

To find the inverse "logarithm with base fraction $\frac{p}{q}$$\frac{p}{q}$ --- "approach the $\mathrm{log}$$\log$ as the equivalent exponent". Logarithm with a fraction as base is not studied.

Consider ${\left(\text{base}\right)}^{\text{power}}=\text{exp. result}$${\left(\textrm{b a s e}\right)}^{\textrm{p o w e r}} = \textrm{\exp . r e s \underline{t}}$

Given that $\text{base}$$\textrm{b a s e}$ is a fraction.
The exponentiation is
${a}^{b}={\left(\frac{p}{q}\right)}^{b}=c=\frac{{p}^{b}}{{q}^{b}}$${a}^{b} = {\left(\frac{p}{q}\right)}^{b} = c = \frac{{p}^{b}}{{q}^{b}}$

The result $c$$c$ is a fraction.
The following are noted.

•  The inverse "root", the numerator and denominator can be independently considered as $\sqrt[b]{c}=\frac{\sqrt[b]{\text{numerator}}}{\sqrt[b]{\text{denominator}}}$$\sqrt[b]{c} = \frac{\sqrt[b]{\textrm{\nu m e r a \to r}}}{\sqrt[b]{\textrm{\mathrm{de} n o \min a \to r}}}$.

•  The inverse "logarithm", ${\mathrm{log}}_{a}c$${\log}_{a} c$, though the base $a$$a$ can be a fraction, it is not formally studied.

If we encounter ${\mathrm{log}}_{a}c$${\log}_{a} c$ where the base $a$$a$ is a fraction, then that is converted into an equivalent equation ${\mathrm{log}}_{a}c={\mathrm{log}}_{b}c÷{\mathrm{log}}_{b}a$${\log}_{a} c = {\log}_{b} c \div {\log}_{b} a$ and solved.

fraction as power

Given that $\text{power}$$\textrm{p o w e r}$ is a positive fraction.

${a}^{b}={a}^{\frac{p}{q}}=c$${a}^{b} = {a}^{\frac{p}{q}} = c$

The inverse : "root" can be
$\sqrt[\frac{p}{q}]{c}=a$$\sqrt[\frac{p}{q}]{c} = a$
${c}^{\frac{q}{p}}=a$${c}^{\frac{q}{p}} = a$

Consider ${\left(\text{base}\right)}^{\text{power}}=\text{exp. result}$${\left(\textrm{b a s e}\right)}^{\textrm{p o w e r}} = \textrm{\exp . r e s \underline{t}}$

Given that $\text{power}$$\textrm{p o w e r}$ is a fraction.
The exponentiation is
${a}^{b}={\left(a\right)}^{\frac{p}{q}}=c=\sqrt[q]{{a}^{p}}$${a}^{b} = {\left(a\right)}^{\frac{p}{q}} = c = \sqrt[q]{{a}^{p}}$

The following are noted.

•  In the inverse "root", $\sqrt[b]{c}$$\sqrt[b]{c}$, the power $b=\frac{p}{q}$$b = \frac{p}{q}$ can be a fraction. The same is considered as ${c}^{\frac{q}{p}}$${c}^{\frac{q}{p}}$.

•  In the inverse "logarithm", ${\mathrm{log}}_{a}c$${\log}_{a} c$, the expression involves base $a$$a$ and value $c$$c$ and the variable under consideration is the result -- which can be a fraction.

base 0

Consider ${\left(\text{base}\right)}^{\text{power}}=\text{exp. result}$${\left(\textrm{b a s e}\right)}^{\textrm{p o w e r}} = \textrm{\exp . r e s \underline{t}}$

Given that $\text{base}$$\textrm{b a s e}$ is $0$$0$ and $\text{power}$$\textrm{p o w e r}$ is a positive-non-zero value.
${0}^{b}=0$${0}^{b} = 0$

The following are noted.

•  $\sqrt[b]{0}=0$$\sqrt[b]{0} = 0$.

•  In ${a}^{b}={0}^{b}=c=0$${a}^{b} = {0}^{b} = c = 0$, the value of $c$$c$ can never be a non-zero value. So $\mathrm{log}$$\log$ to the base $0$$0$ is not defined for any non-zero values .

•  In Further more, ${a}^{b}={0}^{b}=c=0$${a}^{b} = {0}^{b} = c = 0$, the value of $b$$b$ can be any number and does not solve to one value (meaning it has many solutions). So ${\mathrm{log}}_{0}0=b$${\log}_{0} 0 = b$ is also not defined.

power 0

Consider ${\left(\text{base}\right)}^{\text{power}}=\text{exp. result}$${\left(\textrm{b a s e}\right)}^{\textrm{p o w e r}} = \textrm{\exp . r e s \underline{t}}$

Given that $\text{base}$$\textrm{b a s e}$ is a positive non-zero value and $\text{power}$$\textrm{p o w e r}$ is $0$$0$.
${a}^{0}=1$${a}^{0} = 1$

The following are noted

•  ${\mathrm{log}}_{a}1=0$${\log}_{a} 1 = 0$

•  In ${a}^{b}={a}^{0}=c=1$${a}^{b} = {a}^{0} = c = 1$, the value of $c$$c$ can never be other than $1$$1$. So $\sqrt{c}$$\sqrt{c}$ is not defined for any value other than $1$$1$.

•  In ${a}^{b}={a}^{0}=c=1$${a}^{b} = {a}^{0} = c = 1$, the value of $a$$a$ can be any number and does not solve to one value (meaning it has many solutions). So, $\sqrt{1}=a$$\sqrt{1} = a$ is also not defined as $a$$a$ can be any number.

base 1

Consider ${\left(\text{base}\right)}^{\text{power}}=\text{exp. result}$${\left(\textrm{b a s e}\right)}^{\textrm{p o w e r}} = \textrm{\exp . r e s \underline{t}}$

Given that $\text{base}$$\textrm{b a s e}$ is $1$$1$
${1}^{b}=1$${1}^{b} = 1$

The following are noted.

•  $\sqrt[b]{1}=1$$\sqrt[b]{1} = 1$. some additional learnings will be introduced in the higher mathematics, complex numbers

•  In ${a}^{b}={1}^{b}=c=1$${a}^{b} = {1}^{b} = c = 1$, the value of $c$$c$ can never be any value other than $1$$1$. So $\mathrm{log}$$\log$ to the base $1$$1$ is not defined for any numbers that is not $1$$1$.

•  In ${a}^{b}={1}^{b}=c=1$${a}^{b} = {1}^{b} = c = 1$, the value of $b$$b$ can be any number and does not solve to one value (meaning it has many solutions). So, ${\mathrm{log}}_{1}1=b$${\log}_{1} 1 = b$ is also not defined as $b$$b$ can be any number.

summary

Consider ${\left(\text{base}\right)}^{\text{power}}=\text{exp. result}$${\left(\textrm{b a s e}\right)}^{\textrm{p o w e r}} = \textrm{\exp . r e s \underline{t}}$ OR ${a}^{b}=c$${a}^{b} = c$

Summary of what we have learned in understanding roots and logarithms.

•  ${\mathrm{log}}_{\text{-ve}\phantom{\rule{1ex}{0ex}}a}c=b$${\log}_{\textrm{- v e} a} c = b$ is not formally studied, instead convert to exponent form ${a}^{b}=c$${a}^{b} = c$ to solve for $b$$b$.

•  $\sqrt[\text{-ve}\phantom{\rule{1ex}{0ex}}b]{c}=a$$\sqrt[\textrm{- v e} b]{c} = a$ is not formally studied, instead convert to exponent form ${a}^{b}=c$${a}^{b} = c$ to solve for $a$$a$.

•  ${\mathrm{log}}_{a=p/q}c=b$${\log}_{a = p / q} c = b$ is not formally studied, instead convert to exponent form ${a}^{b}=c$${a}^{b} = c$ to solve for $b$$b$.

•  ${\mathrm{log}}_{\text{+ve}\phantom{\rule{1ex}{0ex}}a}\left(\text{-ve}\phantom{\rule{1ex}{0ex}}c\right)=b$${\log}_{\textrm{+ v e} a} \left(\textrm{- v e} c\right) = b$ is not defined, as no value of $b$$b$ satisfies ${\left(\text{+ve}\phantom{\rule{1ex}{0ex}}a\right)}^{b}=\text{-ve}\phantom{\rule{1ex}{0ex}}c$${\left(\textrm{+ v e} a\right)}^{b} = \textrm{- v e} c$.

•  $\sqrt{c}=a$$\sqrt{c} = a$ is not defined, as no value of $a$$a$ satisfies ${a}^{0}=c$${a}^{0} = c$ for $c\ne 1$$c \ne 1$ and $a$$a$ takes many possible values if $c=1$$c = 1$

•  ${\mathrm{log}}_{0}c=b$${\log}_{0} c = b$ is not defined, as no value of $b$$b$ satisfies ${0}^{b}=c$${0}^{b} = c$ for $c\ne 0$$c \ne 0$ and $b$$b$ takes many possible values if $c=0$$c = 0$.

•  ${\mathrm{log}}_{1}c=b$${\log}_{1} c = b$ is not defined, as no value of $b$$b$ satisfies ${1}^{b}=c$${1}^{b} = c$ for $c\ne 1$$c \ne 1$ and $b$$b$ takes many possible values if $c=1$$c = 1$.

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