# Logarithms

Emma was working on a problem related to exponents and for one of the problems, she found that $$2^5 = 32$$

She then thought that while $$2^5$$ can be easily solved by fining the successive exponents of 2, how would she solve a problem that had the question $$2^x = 30^y$$

Is there a way to solve it?

Emma then approached her teacher, who told her to use a logarithm table.

Logarithms or logs are another way of writing exponents and can be used to solve problems which cannot be solved using the concept of exponents only.

In this chapter, we will understand logs and know its meaning, as well as use a log calculator to solve problems.

Check out the interactive simulations to know more about the lesson and try your hand at solving a few interesting practice questions at the end of the page.

## Lesson Plan

 1 What Are Logs? 2 Important Notes on Logs 3 Tips and Tricks 4 Solved Examples on Logs 5 Interactive Questions on Logs

## What Are Logs?

### Logs Meaning

Logs (or) logarithms are nothing but another way of expressing exponents.

Understanding logs is not so difficult.

To understand logs, it is sufficient to know that a logarithmic equation is just another way of writing an exponential equation.

Here is the mathematical definition of logs.

### Logs - Definition

A logarithm is defined using an exponent.

 $$b^x = a \Leftrightarrow \log_b a = x$$

The right side part of the arrow is read to be "Logarithm of $$a$$ to the base $$b$$ is equal to $$x$$".

Here,

• $$a$$ and $$b$$ are two positive real numbers.
• $$x$$ is a real number.
• $$a$$, which is inside the log is called the "argument".
• $$b$$, which is at the bottom of the log is called the "base".

Here is the simulation to understand how an exponent is converted into a logarithm.

Examples of Logs

Exponent

Logarithm

$2^5=32$

$\log_2 32=5$

$6^2 = 36$

$\log_6 36=2$

$3^{-2} = \dfrac{1}{9}$

$\log_3 \dfrac{1}{9}=-2$

## What Are Natural and Common Logs?

### Natural Logs

Natural logs are nothing but logs with base $$e$$.

That is, a natural log means $$\log_e$$.

But it is not usually represented as $$\log_e$$. Instead, it is represented as $$\ln$$.

 $$\log_e = \ln$$

Example

$e^x = m \Rightarrow \log_e m = x \Rightarrow \ln m = x$

### Common Logs

Common logs are nothing but logs with base 10

That is, a common log means $$\log_{10}$$

But usually, writing $$\log$$ is sufficient instead of writing $$\log_{10}$$

 $$\log_{10} = \log$$

It means we write the base of a logarithm only if it is some number other than 10

If there is no base to a logarithm, it means that the base is 10

Example

\begin{align} 7^3 &=343 \Rightarrow \log_7 343 = 3 \\[0.3cm] 10^2 &=100\Rightarrow \log 100= 2 \end{align}

Have you observed that we have not written 10 as the base in the last example, because that's understood?

Important Notes
1. An exponential equation is converted into a logarithmic equation using:
$b^x = a \Leftrightarrow \log_b a = x$
Here, the base of the exponential equation is the same as that of the logarithmic equation.
2. A common log is a logarithm with base 10, i.e.
$\log_{10} = \log$
3. A natural log is a logarithm with base $$e$$, i.e.
$\log_e = \ln$

## Rules (or) Properties of Logs

Here are the rules/properties of logarithms.

### Product Rule

The logarithm of a product of two numbers is the sum of the logarithms of the individual numbers, i.e.

 $$\log_b (mn) = \log_b m + \log_b n$$

Note that the bases of all logs must be the same here.

This resembles/is derived from the product rule of exponents:

$x^m \cdot x^n = x^{m+n}$

### Quotient Rule

The logarithm of a quotient of two numbers is the difference between the logarithms of the individual numbers, i.e.

 $$\log_b \left( \dfrac{m}{n} \right) = \log_b m - \log_b n$$

Note that the bases of all logs must be the same here.

This resembles/is derived from the quotient rule of exponents:

$\dfrac{x^m}{x^n}= x^{m-n}$

### Power Rule

The exponent of the argument of a logarithm can be brought in front of the logarithm, i.e.

 $$\log_b a^m = m \log_b a$$

Note that the bases of all logs must be the same here.

This resembles/is derived from the power of power rule of exponents:

$(x^m)^n = x^{mn}$

### Change of Base Rule

The base of a logarithm can be changed using this property.

 $$\log _{b} a=\dfrac{\log _{c} a}{\log _{c} b}$$

Another way of writing this rule is:

 $$\log _{b} a \, \cdot\, \log _{c} b=\log _{c} a$$

Using this property, we can change the base to any other number.

Hence we can change the base to 10 as well. Then we get:

$\log_b a = \dfrac{\log a}{\log b}$

This equation is really helpful while calculating the logarithm of a number using a calculator because we do not have a button on the calculator to calculate the logarithm of different bases other than 10.

You can see the usage of this rule in the following simulation which is a "Log Calculator".

### Equality Rule

This rule is used while solving the equations involving logarithms.

 $$\log _{b} a= \log_b c \Rightarrow a=c$$

### Other Rules

We apply the definition of logs to some rules/facts of exponents to derive some other rules of logs.

\begin{align} a^0=1 & \Rightarrow \log_a 1 =0 \\[0.2cm] 10^0=1 & \Rightarrow \log 1 =0 \\[0.2cm] e^0=1 & \Rightarrow \ln 1 =0 \\[0.2cm] a^1=a & \Rightarrow \log_a a = 1 \end{align}

Thus the other rules are:

 \begin{align} \log_a 1& =0 \0.2cm] \log 1 &=0 \\[0.2cm] \ln 1 &=0 \\[0.2cm] \log_a a &= 1\\[0.2cm] a^{\log_a x}&=x\\[0.2cm]\log_{b^n}a^m&= \dfrac{m}{n}\log_b a \end{align} Tips and Tricks 1. The logarithm of 0 is NOT defined as one number raised to another number never gives 0 as the result. 2. The above-mentioned properties hold even for common and natural logarithms. For example, from the product rule, \[ \begin{align}\ln (mn) &= \ln m + \ln n \end{align}
3. By the change of base property, $$\log_b a= \dfrac{1}{\log_a b}$$

## What Are Negative Logs?

The negative logs are of the form $$-\log_b a$$.

We can calculate this using the power rule of logarithms:

\begin{align} - \log_b a &= \log_b a^{-1}\\[0.2cm] &= \log_b \dfrac{1}{a} \end{align}

Thus,

 $$- \log_b a = \log_b \dfrac{1}{a}$$

To convert a negative log into a positive log, we can just take the reciprocal of the argument.

Also, to convert a negative log into a positive log, we can take the reciprocal of the base, i.e.

 $$- \log_b a = \log_\dfrac{1}{b} a$$

## How to Expand Logs?

How to do logs?

We just apply the rules of logs to expand logs.

Example:

\begin{align} &\log (3x^2y^3)\\[0.2cm]&= \log (3) + \log (x^2)+ \log (y^3) \,\, \\[0.2cm] & [ \text{By product rule}]\\[0.2cm] &= \log 3 + 2 \log x + 3 \log y \,\, \\[0.2cm] & [ \text{By power rule}] \end{align}

## Solved Examples

 Example 1

Patrick is instructed to find the value of the logarithm $$\log _{5} \frac{1}{25}$$. He has been told that the answer must be an integer.

Solution

We will help Patrick to convert the given logarithm into an integer.

\begin{align} \log _{5} \frac{1}{25} &= \log_5 1 - \log_5 25\\[0.2cm] &= 0 - \log_5 5^2\\[0.2cm] &= -\log_5 5^2\\[0.2cm] &= -2 \log_5 5\\[0.2cm] &= -2 (1)\\[0.2cm] &=-2\end{align}

 $$\therefore \log _{5} \frac{1}{25} = -2$$
 Example 2

Jerry is trying to expand the following logarithm. Shall we help him?

$\log _{6}\left(\frac{6 \mathrm{m}^{3}}{\sqrt{\mathrm{n}}}\right)$

Solution

We will help Jerry to expand the given logarithm using the properties (or) rules of logarithms:

\begin{align} &\log _{6}\left(\frac{6 \mathrm{m}^{3}}{\sqrt{\mathrm{n}}}\right) \\ &=\log _{6}\left(6 \mathrm{m}^{3}\right)-\log _{6}(\sqrt{\mathrm{n}}) \\[0.2cm] &=\log _{6}(6)+\log _{6}\left(\mathrm{m}^{3}\right)-\log _{6}\left(\mathrm{n}^{\frac{1}{2}}\right) \\[0.2cm] &=1+3 \log _{6}(\mathrm{m})-\frac{1}{2} \log _{6}(\mathrm{n}) \end{align}

Thus, the expanded form of the given logarithm is

 $$\therefore 1+3 \log _{6}(\mathrm{m})-\frac{1}{2} \log _{6}(\mathrm{n})$$
 Example 3

Compress the following expression into a single logarithm using the properties of logarithms:

$5 \log _{3}(x)+2 \log _{3}(4 x)-\log _{3}\left(8 x^{5}\right)$

Solution

We will compress the given expression using the properties of logarithms:

\begin{align}&5 \log _{3}(x)+2 \log _{3}(4 x)-\log _{3}\left(8 x^{5}\right)\\ &=\log _{3}\left(x^{5}\right)+\log _{3}\left( (4 x)^{2}\right)-\log _{3}\left(8 x^{5}\right) \0.3cm] &=\log _{3}\left(x^{5}\right)+\log _{3}\left(16 x^{2}\right)-\log _{3}\left(8 x^{5}\right) \\[0.3cm] &=\log _{3}\left(x^{5} \cdot 16 x^{2}\right)-\log _{3}\left(8 x^{5}\right) \\[0.3cm] &=\log _{3}\left(16 x^{7}\right)-\log _{3}\left(8 x^{5}\right) \\[0.3cm] &=\log _{3}\left(\frac{16 x^{7}}{8 x^{5}}\right)\\[0.3cm] &= \log _{3}\left(2 x^{2}\right) \end{align}  $$\therefore \log _{3}\left(2 x^{2}\right)$$  Example 4 Mia invested 1000 in a trust fund at an annual percentage rate of 12% compounded yearly. How long will it take the investment to reach 2000? Round your answer to the nearest tenths. Solution The initial investment is $$P= \ \, 1000$$. The rate of interest is $$r=12\% =0.12$$. The final amount is $$A= \ \, 2000$$. Let us assume the required time to be $$t$$ years. Since the amount is compounded yearly, $$n=1$$. We will substitute all these values in the compound interest formula: \[\begin{align} A &= P \left(1+ \dfrac{r}{n} \right)^{nt}\\[0.2cm] 2000&=1000 \left(1+ \dfrac{0.12}{1} \right)^{1 \cdot t}\\[0.2cm] 2 &= (1.12)^t\\[0.2cm] t&= \log_{1.12} 2\\[0.2cm] t&= \dfrac{\log 2}{\log 1.12}\\[0.2cm] t& \approx 6.1 \end{align}

 $$\therefore$$ the required time is $$\ 6.1\ years$$

## Interactive Questions

Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.

## Let's Summarize

We hope you enjoyed learning about Logs with the simulations and practice questions. Now you will be able to easily solve problems on logs meaning, understanding logs, log calculator, and how to do logs.

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

## 1. Why do we use logs?

We use logs to solve problems involving exponents.

## 2. How do you simplify logs?

We can simplify logs using the rule (or) properties of logs.

## 3. Is log 0 possible?

The logarithm of 0 is NOT defined as one number raised to another number never gives 0 as the result.

Think whether $$a^b=0$$ for any values of $$a$$ and $$b$$.

Exponents and Logarithms
Grade 9 | Questions Set 1
Exponents and Logarithms
Exponents and Logarithms
Grade 9 | Questions Set 2
Exponents and Logarithms