Exponents : Understanding exponents, roots, logarithms

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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 ( $\sqrt[- 2]{4}$ $root(-2)(4)$ )

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

• logarithm with fraction base ( $\log_{\frac{1}{2}} 8$ $log_(1/2) 8$ )

• logarithm of negative value ( $\log_{2} (- 4)$ $log_2 (-4)$ )

• logarithm with negative base ( $\log_{- 2} (- 8)$ $log_(-2) (-8)$ )

• logarithm with 0 base ( $\log_{0} 2$ $log_0 2$ )

• logarithm with 1 base ( $\log_{1} 2$ $log_1 2$ )

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

If exponentiation result and power are given, then
$base = \sqrt[power]{result}$ $text(base) = root(text(power))(text(result))$
This inverse is called the "root".

If exponentiation-result and base are given, then
$power = \log_{base} (exp.result)$ $text(power) = log_text(base) (text(exp.result))$
This inverse is called the "logarithm".

all combinations and their uses

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

$base$ $text(base)$ and $power$ $text(power)$ 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 ${(base)}^{power} = exp. result$ $(text(base))^(text(power)) = text(exp. result)$

Given that $base$ $text(base)$ and $power$ $text(power)$ are positive integers. The $exp. result$ $text(exp. result)$ a positive integer".

negative base - not so simple

Given that, $base$ $text(base)$ is a negative integer and $power$ $text(power)$ is a positive integer. The $exp. result$ $text(exp. result)$ 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
${(- a)}^{b} = d$ $(-a)^b = d$
The result $d$ $d$ can be a positive or a negative value.

The equation can be simplified to
${(- a)}^{b} = {(- 1)}^{b} a^{b} = {(- 1)}^{b} c$ $(-a)^b = (-1)^b a^b = (-1)^b c$
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. $\log$ $log$ of negative values are not defined.

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

Note 2: If we encounter $\log_{-ve a} (+ve c) = b$ $log_(text(-ve ) a) (text(+ve ) c) = b$ or $\log_{-ve a} (-ve c) = b$ $log_(text(-ve ) a) (text(-ve ) c) = b$ , then convert them into the equivalent exponent equation $a^{b} = c$ $a^b=c$ and solve for $b$ $b$ .

negative power

Given that $base$ $text(base)$ is a positive integer and $power$ $text(power)$ is a negative integer. $exp. result$ $text(exp. result)$ is "a positive fraction".

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

So, the result is a positive fraction.

Given that $base$ $text(base)$ is a positive integer and $power$ $text(power)$ is a negative integer.

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

The exponentiation is
$a^{- b} = \frac{1}{a^{b}} = {(\frac{1}{a})}^{b} = c = \frac{1}{d}$ $a^(-b) = 1/(a^b) = (1/a)^b = c = 1/d$
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 $\frac{1}{a}$ $1/a$ . This is explained along with the generic form of base $\frac{p}{q}$ $p/q$ in the next page.

fraction as base

Given that $base$ $text(base)$ is a positive fraction.
$a^{b} = {(\frac{p}{q})}^{b} = c$ $a^b = (p/q)^(b) =c$

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

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

Given that $base$ $text(base)$ is a fraction.
The exponentiation is
$a^{b} = {(\frac{p}{q})}^{b} = c = \frac{p^{b}}{q^{b}}$ $a^b = (p/q)^(b) =c = (p^b)/(q^b)$

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

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

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

If we encounter $\log_{a} c$ $log_a c$ where the base $a$ $a$ is a fraction, then that is converted into an equivalent equation $\log_{a} c = \log_{b} c \div \log_{b} a$ $log_a c = log_b c -: log_b a$ and solved.

fraction as power

Given that $power$ $text(power)$ is a positive fraction.

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

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

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

Given that $power$ $text(power)$ is a fraction.
The exponentiation is
$a^{b} = {(a)}^{\frac{p}{q}} = c = \sqrt[q]{a^{p}}$ $a^b = (a)^(p/q) =c = root(q)(a^p)$

The following are noted.

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

• In the inverse "logarithm", $\log_{a} c$ $log_a c$ , 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 ${(base)}^{power} = exp. result$ $(text(base))^(text(power)) = text(exp. result)$

Given that $base$ $text(base)$ is $0$ $0$ and $power$ $text(power)$ is a positive-non-zero value.
$0^{b} = 0$ $0^b = 0$

The following are noted.

• $\sqrt[b]{0} = 0$ $root(b)(0) = 0$ .

• 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 $\log$ $log$ 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 $\log_{0} 0 = b$ $log_0 0 = b$ is also not defined.

power 0

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

Given that $base$ $text(base)$ is a positive non-zero value and $power$ $text(power)$ is $0$ $0$ .
$a^{0} = 1$ $a^0 = 1$

The following are noted

• $\log_{a} 1 = 0$ $log_a 1 = 0$

• 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 $\sqrt[0]{c}$ $root(0)(c)$ 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, $\sqrt[0]{1} = a$ $root(0)(1) = a$ is also not defined as $a$ $a$ can be any number.

base 1

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

Given that $base$ $text(base)$ is $1$ $1$
$1^{b} = 1$ $1^b = 1$

The following are noted.

• $\sqrt[b]{1} = 1$ $root(b)(1) = 1$ . 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 $\log$ $log$ 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, $\log_{1} 1 = b$ $log_1 1 = b$ is also not defined as $b$ $b$ can be any number.

summary

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

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

• $\log_{-ve a} c = b$ $log_(text(-ve ) a) c = b$ is not formally studied, instead convert to exponent form $a^{b} = c$ $a^b=c$ to solve for $b$ $b$ .

• $\sqrt[-ve b]{c} = a$ $root(text(-ve ) b)(c) = a$ is not formally studied, instead convert to exponent form $a^{b} = c$ $a^b=c$ to solve for $a$ $a$ .

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

• $\log_{+ve a} (-ve c) = b$ $log_(text(+ve ) a) (text(-ve ) c) = b$ is not defined, as no value of $b$ $b$ satisfies ${(+ve a)}^{b} = -ve c$ $(text(+ve ) a)^b = text(-ve ) c$ .

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

• $\log_{0} c = b$ $log_0 c = b$ is not defined, as no value of $b$ $b$ satisfies $0^{b} = c$ $0^b = c$ for $c \neq 0$ $c!=0$ and $b$ $b$ takes many possible values if $c = 0$ $c=0$ .

• $\log_{1} c = b$ $log_1 c = b$ is not defined, as no value of $b$ $b$ satisfies $1^{b} = c$ $1^b = c$ for $c \neq 1$ $c!=1$ 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