overview

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

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

• root with power 0 ($\sqrt[0]{4}$)

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

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

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

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

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

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

$\left(\text{base}\right)}^{\text{power}}=\text{exp. result$

If exponentiation result and power are given, then

$\text{base}=\sqrt[\text{power}]{\text{result}}$

This inverse is called the "root".

If exponentiation-result and base are given, then

$\text{power}={\mathrm{log}}_{\text{base}}\left(\text{exp.result}\right)$

This inverse is called the "logarithm".

all combinations and their uses

Consider $\left(\text{base}\right)}^{\text{power}}=\text{exp. result$

$\text{base}$ and $\text{power}$ can be one of positive integers, negative integers, $0$, $1$, 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 $\left(\text{base}\right)}^{\text{power}}=\text{exp. result$*

Given that $\text{base}$ and $\text{power}$ are positive integers. The $\text{exp. result}$ a positive integer".

negative base - not so simple

Given that, $\text{base}$ is a negative integer and $\text{power}$ is a positive integer. The $\text{exp. result}$ can be a positive or negative value".

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

The exponentiation is

${(-a)}^{b}=d$

The result $d$ can be a positive or a negative value.

The equation can be simplified to

${(-a)}^{b}={(-1)}^{b}{a}^{b}={(-1)}^{b}c$

and so, *the value $c$ 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 $c$) 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 $a$) is a positive number. Negative bases are not studied, as the negative base is handled as power of $-1$ as shown.

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

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

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

negative power

Given that $\text{base}$ is a positive integer and $\text{power}$ is a negative integer. $\text{exp. result}$ is "a positive fraction".

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

So, the result is a positive fraction.

Given that $\text{base}$ is a positive integer and $\text{power}$ is a negative integer.

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

The exponentiation is

$a}^{-b}=\frac{1}{{a}^{b}}={\left(\frac{1}{a}\right)}^{b}=c=\frac{1}{d$

The result $c$ 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 ${a}^{b}=c$ and solve for the unknown value.

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

fraction as base

Given that $\text{base}$ is a positive fraction.

${a}^{b}={\left(\frac{p}{q}\right)}^{b}=c$

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

*Consider $\left(\text{base}\right)}^{\text{power}}=\text{exp. result$*

Given that $\text{base}$ is a fraction.

The exponentiation is

$a}^{b}={\left(\frac{p}{q}\right)}^{b}=c=\frac{{p}^{b}}{{q}^{b}$

The result $c$ is a fraction.

The following are noted.

• The inverse "root", the numerator and denominator can be independently considered as $\sqrt[b]{c}=\frac{\sqrt[b]{\text{numerator}}}{\sqrt[b]{\text{denominator}}}$.

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

If we encounter ${\mathrm{log}}_{a}c$ where the base $a$ is a fraction, then that is converted into an equivalent equation ${\mathrm{log}}_{a}c={\mathrm{log}}_{b}c\xf7{\mathrm{log}}_{b}a$ and solved.

fraction as power

Given that $\text{power}$ is a positive fraction.

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

The inverse : "root" can be

$\sqrt[\frac{p}{q}]{c}=a$

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

*Consider $\left(\text{base}\right)}^{\text{power}}=\text{exp. result$*

Given that $\text{power}$ is a fraction.

The exponentiation is

$a}^{b}={\left(a\right)}^{\frac{p}{q}}=c=\sqrt[q]{{a}^{p}$

The following are noted.

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

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

base 0

*Consider $\left(\text{base}\right)}^{\text{power}}=\text{exp. result$*

Given that $\text{base}$ is $0$ and $\text{power}$ is a positive-non-zero value.

${0}^{b}=0$

The following are noted.

• $\sqrt[b]{0}=0$.

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

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

power 0

*Consider $\left(\text{base}\right)}^{\text{power}}=\text{exp. result$*

Given that $\text{base}$ is a positive non-zero value and $\text{power}$ is $0$.

${a}^{0}=1$

The following are noted

• ${\mathrm{log}}_{a}1=0$

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

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

base 1

*Consider $\left(\text{base}\right)}^{\text{power}}=\text{exp. result$*

Given that $\text{base}$ is $1$

${1}^{b}=1$

The following are noted.

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

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

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

summary

*Consider $\left(\text{base}\right)}^{\text{power}}=\text{exp. result$ OR ${a}^{b}=c$*

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

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

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

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

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

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

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

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

Outline

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

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