# Logarithms

A **logarithm** is the inverse of the exponential function. Specifically, the logarithm returns the value of the exponent required to raise the base to the specified value.

For example, the base-2 logarithm of \(64\) is \(6,\) because \( 2^6 = 64.\) In general, we have the following:

\( z \) is the

base-\(x\) logarithm of \(y\)if and only if \( x^z = y \). In typical notation\[ \log_x y = z \iff x^z = y.\]

#### Contents

## Properties of Logarithms - Basic

First, we must know the basic structure of a logarithm \((\)abbreviated \(\log\) for convenience\().\) \(\log_a{b}=c\) can be rewritten as \(a^c=b,\) where \(a\) is called the **base**, \(c\) the **exponent**, and \(b\) the **argument**. Also, \(\log\) without a base is shorthand for the common \(\log\) of base \(10.\) Now that we know this, we can manipulate logs:

In Math | In English | Example |

\(\log_a b + \log_a c = \log_a bc\) | When you add logs with the same base, you can merge into one log and multiply their arguments. | \(\log_2 5 + \log_2 6 = \log_2 30\) |

\(\log_a{b}-\log_a{c} = \log_a \frac{b}{c}\) | The opposite of the above. | \(\log_2{18}-\log_2{6} = \log_2{3}\) |

\(\log_a b^c = c \cdot \log_a b\) | When your result has an exponent, you can move it to the front of the log. | \(\log_3{\big(5^{-4}\big)} = -4 \cdot \log_3{5}\) |

\(\log_{a^b}{c}=\frac{1}{b}\log_{a}{c}\) | When the base has an exponent, you can move its reciprocal to the front of the log. | \(\log_{\pi^2}{e}=\frac{1}{2}\log_{\pi}{e}\) |

\(\log_a b = \frac{\log_c b}{\log_c a}\) | You can rearrange any log by making a fraction, with log of the argument in the numerator and the log of the base in the denominator. Any base can be chosen for the logs, but the bases must be the same for both logs. | \(\log_2 \pi = \frac{\log \pi}{\log 2}\) |

\(log_a b=\frac{1}{\log_b a}\) | If you want to switch the base of the log \(a\) with the argument \(b,\) then you take the reciprocal. | \(\log_4 e=\frac{\log_e e}{\log_e 4}=\frac{1}{\log_e 4}\) |

\({ a }^{ \log _{ a }{ b } } = b\) | When a constant \(a\) is raised to the power \(\log _{ a }{ b },\) the resultant expression is \(b.\) | \({ e }^{ \log _{ e }{ 3 } }=3\) |

\(\log_{a}{1}=0\) | Any log which has 1 as its argument will be equal to 0. | \(\log_{\pi \cdot e}{1}=0\) |

Other properties can be derived from these basic ones, especially when noting that these properties are inversable.

## Simplify \(\log_2 \left(\dfrac{32}{9}\right)^2\) as much as possible.

Try to follow the steps and identify what properties were used:

\[\begin{align} \log_2 \left(\dfrac{32}{9}\right)^2 &=2 \cdot \log_2 \left(\frac{32}{9}\right)\\ &=2 \cdot ( \log_2 32 - \log_2 9)\\ &=2 \cdot \left( \log_2 2^5 -\log_2 3^2\right)\\ &=2 \cdot ( 5 \cdot \log_2 2 - 2 \cdot \log_2 3)\\ &=2 \cdot ( 5 \cdot 1 -2 \cdot \log_2 3)\\ &=10- 4 \log_2 3. \end{align}\]

Note: \(\log_2 3\) can't be simplified further. Line 1 used the second property, line 2 put thingies into exponential form, line 3 used the third property, and lines 4 and 5 did basic simplification. \(_\square\)

## Simplify \[\displaystyle{2\log_4{\sqrt{5}}+\frac{1}{2}\log_2{625}-\log_2{\frac{1}{5}}}.\]

Again, try to follow the steps of the solution:

\[\begin{align} 2\log_{(2^2)}{\big(5^\frac{1}{2}\big)}+\frac{1}{2}\log_2{\big(5^4\big)}-\log_2{\big(5^{-1}\big)} &=2\frac{\log 5^{\frac{1}{2}}}{\log 2^2}+\frac{1}{2}(4)(\log_2{5})+\log_2{5}\\ &=2\frac{\frac{1}{2}\log 5}{2\log 2}+2\log_2{5}+\log_2{5}\\ &=\frac{1}{2}\frac{\log 5}{\log 2}+3\log_2{5}\\ &=\frac{1}{2}\log_2{5}+3\log_2{5}\\ &=\frac{7}{2}\log_2{5}. \end{align}\]

The first line shows that it is (usually) best to convert numbers so that they are integers to a power. Note that lines 4 reverses the process of the fourth property. \(_\square\)

## Worked Examples Using Properties

\[ \] \(1.~\log _{ a }{ a } =1\)

## Find the value of \(\log _{ 4 }{ 4 }.\)

Using the property \(\log _{ a }{ a }=1,\) we get \(\log_{ 4 }{ 4 } =1. \ _\square\)

\[ \] \(2.~\log _{ a }{ (b^c) } =c\log _{ a }{ b } \)

## Find the value of \(\log _{ 2 }{ 16 }.\)

We have

\[\begin{align} \log _{ 2 }{ 16 } &= \log _{ 2 }{ { 2 }^{ 4 } } &&\qquad \big(16={ 2 }^{ 4 }\big)\\ &=4\log _{ 2 }{ 2 } &&\qquad \big(\log { { a }^{ b } } = b\log { a } \big)\\ &= 4. \ _\square &&\qquad (\text{by property 1}) \end{align}\]

\[ \] \(3.~\log _{ a }{ (b \times c) } = \log _{ a }{ b }+ \log _{ a }{ c } \)

## Find the value of \(\log { 90 }\) assuming \(\log { 3 } =0.47\).

We have

\[\begin{align} \log { 90 } &= \log { (9\times 10) } &&\qquad (90= 9 \times 10)\\ &=\log { 9 } + \log { 10 } &&\qquad \big(\log _{ a }{ (b\times c) } =\log _{ a }{ b } +\log _{ a }{ c } \big)\\ &=2\log { 3 } +1 &&\qquad \text{(by properties 2 and 1)}\\ &=2\times 0.47+1\\ &=0.94+1\\ &=1.94. \ _\square \end{align}\]

\[ \] \(4.~\displaystyle{\log _{ a }{ \frac { b }{ c } } = \log _{ a }{ b } - \log _{ a }{ c }}\)

## Evaluate \(\log { 0.27 }\) assuming \(\log { 3 } =0.47\).

We have

\[\begin{align} \log { 0.27 } &= \log { \frac { 27 }{ 100 } } \\ &= \log { 27 } - \log { 100 } \\ &=3\log{ 3 } - 2\\ &=1.41 - 2\\ &=-0.59. \ _\square \end{align}\]

\[ \] \(5.\) \(\displaystyle{\log_{ a }{ b } = \frac { \log_{ c }{ b } }{ \log_{ c }{ a } }} \)

## Find the value of \(\log_{ 32 }{ 2 }\).

We have

\[\begin{align} \log_{ 32 }{ 2 } &=\frac { \log_{ 2 }{ 2 } }{ \log_{ 2 }{ 32 } } \\ &=\frac { 1 }{ 5\log_{ 2 }{ 2 } } \\ &=\frac { 1 }{ 5 }\\ &={ 0.2 }. \ _\square \end{align} \]

## Properties of Logarithms - Intermediate

## What is the value of \( \log_3 15 + \log_3 81 - \log_3 5 ?\)

Using the properties of logarithms, we can rewrite the given expression as follows:

\[ \begin{align} \log_3 15 + \log_3 81 - \log_3 5 &= \log_3 15 - \log_3 5 + \log_3 3^4 \\ &= \log_3 \frac{15}{5} + \log_3 3^4 \\ &= \log_3 3+ 4 \log_3 3 \\ &= 5. \ _\square \end{align}\]

## Problem Solving - Basic

## What is(are) the solution(s) of the quadratic equation

\[\log 2x + \log(x-1) = \log\big(x^2+3\big) ?\]

We have

\[ \begin{align} \log 2x + \log(x-1) &= \log(x^2+3) \\ \log 2x(x-1) &= \log (x^2+3) \\ \Rightarrow 2x(x-1) &= x^2 +3 \\ x^2-2x-3 &= 0 \\ (x+1)(x-3) &= 0 \\ x &= -1, 3. \end{align} \]

Since the logarithm functions \( \log(x-1)\) and \( \log 2x\) are defined over positive numbers, it must be true that \(x-1>0 \implies x>1\) and \(2x>0 \implies x>0.\) Thus, \(-1\) is can not be the value of \(x,\) implying that the value of \(x\) satisfying the given equation is \(x=3.\) \(_\square\)

## What is the solution(s) of the quadratic equation

\[ 2(\log x)^2 = 7\log x - 3 ?\]

We have

\[\begin{align} 2(\log x)^2 &= 7\log x - 3 \\ 2(\log x)^2 - 7\log x +3 &= 0 \\ (\log x -3)(2\log x -1) &= 0 \\ \Rightarrow \log x &= 3, \frac{1}{2} \\ x &= 1000, \sqrt{10}. \ _\square \end{align} \]

## Problem Solving - Intermediate

## If the solutions of the quadratic equation \( x^{\log_3 x\,-\,2} = 27 \) are \(a\) and \(b,\) what is \( \log_{a} b + \log_{b} a?\)

Taking logs with base 3 on both sides, we have

\[ \begin{align} x^{\log_{3} x\,-\,2} &= 27 \\ \Rightarrow (\log_{3} x -2)\log_{3} x &= \log_{3} 27 \\ (\log_{3} x)^2-2\log_{3} x - 3 &= 0 \\ (\log_{3} x +1)(\log_{3} x - 3) &= 0 \\ \log_{3} x &= -1, 3. \end{align} \]

Since \( \log_{a} b + \log_{b} a \) can be expressed as \(\frac{\log_{3} b}{\log_{3} a} + \frac{\log_{3} a}{\log_{3} b}\) using log with base 3,

\[ \begin{align} \log_{a} b + \log_{b} a &= \frac{\log_{3} b}{\log_{3} a} + \frac{\log_{3} a}{\log_{3} b} \\ &= \frac{-1}{3} + \frac{3}{-1} \\ &= -\frac{10}{3}. \ _\square \end{align} \]

If the solutions of the equation \(\log_{2} x + a\log_{x} 8 = b \) are \(2\) and \(\frac{1}{8},\) what are \(a\) and \(b?\)

We have

\[ \begin{align} \log_{2} x + a\log_{x} 8 &= b \\ \log_{2} x + a\log_{x} 2^3 &= b \\ \log_{2} x + \frac{3a}{\log_{2} x} &= b \\ (\log_{2} x)^2 -b \log_{2} x + 3a &= 0. \qquad (1) \\ \end{align} \]

Since the solutions of the equation \( {(\log_{2} x)}^2 -b \log_{2} x + 3a = 0 \) are \(2\) and \(\frac{1}{8} ,\) substituting \(2\) and \(\frac{1}{8}\) into \((1)\) gives

\[ \begin{align} (\log_{2} 2)^2 - b \log_{2} 2 + 3a &= 0 \\ \Rightarrow 1-b+3a &= 0, \qquad (2)\\ (\log_{2} \frac{1}{8} )^2 - b \log_{2} \frac{1}{8} + 3a &= 0 \\ \Rightarrow 9+3b+3a &=0. \qquad (3) \end{align} \]

Solving the simultaneous equations \((2)\) and \((3)\) gives \(a= -1\) and \(b = -2.\) \(_\square\)

## Problem Solving - Advanced

What are the solutions of the equation

\[ \log_{x} xy \times \log_{y} xy + \log_{x}(x-y) \times \log_{y} (x-y) = 0?\]

We have

\[ \begin{align} \log_{x} xy \times \log_{y} xy + \log_{x}(x-y) \times \log_{y} (x-y) &= 0 \\ \frac{\log xy}{\log x} \times \frac{\log xy}{\log y} + \frac{\log(x-y)}{\log x} \times \frac{\log(x-y)}{\log y} &= 0 \\ \frac{(\log xy)^2 + (\log(x-y))^2}{\log x \cdot \log y} &= 0 \\ (\log xy)^2 + (\log(x-y))^2 &= 0 \\ \Rightarrow \log xy &= 0 \text{ and } \log(x-y)= 0. \\ \end{align} \]

Since \(x\) and \(y\) are both positive, this implies that

\[ \begin{align} xy &= 1 \text{ and } x-y=1 \\ \Rightarrow x&= \frac{\sqrt{5} +1}{2}, y=\frac{\sqrt{5}-1}{2}. \ _\square \end{align} \]

## Applications

\[ \]
**Richter Scale:**

Richter scale was developed by Charles Richter in 1935 to compare the intensities of earthquakes. The amount of energy released in an earthquake is very large, so a logarithmic scale avoids the use of large numbers.

The formula used for these calculations is

\[M= \log_{10}\left(\frac{I}{I_0}\right),\]

where \(M\) is the magnitude on the Richter scale, \(I\) is the intensity of the earthquake being measured, and \(I_0\) is the intensity of a reference earthquake.

Let's do a quick example to clarify how this works.

The 1906 San Francisco earthquake had a magnitude of 8.3 on the Richter scale. At the same time in South America there was an eathquake with magnitude 4.1 that caused only minor damage. How many times more intense was the San Francisco earthquake than the South American one?

Because the magnitude is a base-10 log, the Richter number is actually the exponent that 10 is raised to in order to calculate the intensity of the earthquake. Thus, the difference in magnitudes of the earthquakes can be calculated as follows:

\[M=\log_{10}\left(\frac{10^{8.3}}{10^{4.1}}\right)=4.2.\]

So, to answer the question, the San Francisco earthquake is more intense than the South American one by about \(10^M \approx 15848.93192\) times!

Note that you can just subtract 4.1 from 8.3 and get the same result. But if your math teachers are like mine, they will want you to use logarithms, and this is how it is done. The reason that subtracting the magnitudes works is because of the exponent rule for dividing exponents with the same base.

\[ \]
**Decibel Scale:**

One decibel is one tenth of one bel, named in honor of Alexander Graham Bell. The bel is rarely used without the deci- prefix, deci- meaning one tenth. The decibel scale is used to calculate the difference in intensity between two sounds:

\[L=10\log_{10}\left(\frac{I}{I_0}\right),\]

where \( L\) is the loudness of the sound measured in decibels, \(I\) is the intensity of the sound being measured, and \(I_0\) is the intensity of the sound at the threshold of hearing which is equal to zero decibels.

\[ \]
**\(\text{pH}\) Scale:**

The \(\text{pH}\) scale was invented in 1910 by Dr. Soren Sorenson, Head of Laboratory at Carlsberg Beer Company. The "H" in \(\text{pH}\) stands for hydrogen and the meaning of the "p" in \(\text{pH}\), although disputed, is generally considered to mean the power of hydrogen. This scale is used to measure the acidity or alkalinity of water or water soluble substances including, but definitely not limited to, soil or rainwater. The \(\text{pH}\) scale ranges from 1 to 14, where seven is a neutral point. Values below 7 indicate acidity with 1 being the most acidic. Values above 7 indicate alkalinity with14 being the most alkaline:

\[\text{pH}=-\log_{10}\ce{[H+]},\]

where \(\text{pH}\) is the \(\text{pH}\) number between \(1\) and \(14\) and \(\ce{[H+]}\) is the concentration of hydrogen ions.