The concept of logarithms was introduced by John Napier in the ({\rm{1}}{{\rm{7}}^{{\rm{th}}}}) century. Properties of Logarithms are used to solve many mathematical equations. The log features are used to compress numerous logarithms into a single logarithm or expand a single logarithm into multiple logarithms.

Many of the scientists, navigators, engineers, etc., made it simple for performing different calculations. So, the logarithms are the opposite of exponentiation. There are many rules called the laws of logarithms. This allows to express and involve the logarithms to rewrite in various ways. In this article, we will discuss everything about the properties of logarithms.

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What is Logarithm?

Definition: The logarithm is defined using the exponent as follows

\({b^x} = a \Leftrightarrow {\log _b}a = x\)

Here, the right side of the arrow is read to be a logarithm of \(a\) to the base \(b\) is equal to \(x\). Here,

1. \(a\) and \(b\) are considered as the two positive real numbers.
2. \(x\) is the real number.
3. \(a\) which is inside the \(\log \), is known as the argument.
4. \(b\) which is at the bottom of the \(\log \), is known as the base.

Examples: If \({6^2} = 36\) and the logarithm will be \({\log _6}36 = 2\)

Laws of Logarithm

Definition: The logarithmic expressions can be written in different ways, and there are a few specific laws known as the laws of logarithms.

That is \({b^y} = a\) which can be written as \({\log _b}a = y\).

So, this can be expressed as the logarithm of the positive actual number \(a\) to the base\(b\), is a positive \(y\).

Examples:

\({10^1} = 10 \Rightarrow {\log _{10}}10 = 1\)
\({10^2} = 100 \Rightarrow {\log _{10}}100 = 2\)
\({10^3} = 1000 \Rightarrow {\log _{10}}1000 = 3\)
\({10^4} = 10000 \Rightarrow {\log _{10}}10000 = 4\)
\({3^2} = 9 \Rightarrow {\log _3}9 = 2\)
\({3^5} = 243 \Rightarrow {\log _3}243 = 5\)

and so on.

Properties of Logarithms

We have specific rules based on which the logarithmic operations can be achieved, which are given below:

1. Product rule
2. Division rule
3. Power rule or the Exponential Rule
4. Change of base rule
5. Base switch rule
6. Derivative of log
7. Integral of log

Now, we will discuss this one by one along with the examples:

Product Rule

The base will remain the same when the sum of the logarithms of two numbers is identical to the product of the logarithms of the numbers.
It is written as \(\log \,a + \log \,b = \log \,ab\)
Example: (a) \({\log _2}5 + {\log _2}4 = {\log _2}(5 \times 4) = {\log _2}20\)
\({\log {10}}6 + {\log {10}}3 = {\log {10}}(6 \times 3) = {\log {10}}18\)
\(\log x + \log y = \log (x \times y) = \log xy\)
\(\log 4x + \log x = \log (4x \times x) = \log 4{x^2}\)

Division Rule

The base will remain the same when the logarithm of the quotient of two numbers is identical to the difference of the logarithms of the same two numbers.
\({\log _b}\left( {\frac{m}{n}} \right) = {\log _b}m – {\log _b}n\)
Example: \({\log _3}\left( {\frac{2}{y}} \right) = {\log _3}(2) – {\log _3}(y)\)

Power Rule

In this rule, the base remaining the same, the logarithm of \(m\) to a rational exponent, equals the exponent times the logarithm of \(m\).
\({\log {\rm{b}}}\left( {{m^n}} \right) = n{\log {\rm{b}}}m\)
Example: \({\log _b}\left( {{2^3}} \right) = 3{\log _b}2\)

Change of Base Rule

In maths calculations, when you involve the logarithm, you need to change the base of the logarithm. This rule allows you to change the bottom of the logarithm.
Suppose we have the logarithm of a number \(m\) concerning the base \(b\). We want to change the base \(b\) to the base \(a\). In this case, we can change the base as follows:
\({\log {\rm{b}}}m = \frac{{{{\log }{\rm{a}}}m}}{{{{\log }_{\rm{a}}}b}}\)
Example:  \({\log {\rm{b}}}2 = \frac{{{{\log }{\rm{a}}}2}}{{{{\log }_{\rm{a}}}b}}\)

Base Switch Rule

\({\log _b}(a) = \frac{1}{{{{\log }_a}(b)}}\)
Example: \({\log _b}8 = \frac{1}{{{{\log }_8}b}}\)

Derivative of a Log

If \(f(x) = y = {\log _b}(x)\) then we write \(y = {\log _b}(x) = {\log _e}x \times {\log _b}e = {\log _e}x \times \frac{1}{{{{\log }_g}b}}\)
\( \Rightarrow y = \frac{{\ln x}}{{\ln b}}\)
So, the derivative of \(f(x) = y\) is given by \(\frac{{dy}}{{dx}} = \frac{1}{{x \times \ln b}}\)
Example: Given, \(f(x) = y = {\log _{10}}(x)\)
Then, \(\frac{{dy}}{{dx}} = \frac{1}{{x \times \ln 10}}\)

Integral of a Log

\(\int {{{\log }_{\rm{b}}}} (x)dx = \int {{{\log }_e}} x \times {\log _b}edx\)
\( = \int {{{\log }_e}} x \times \frac{1}{{{{\log }_e}b}}dx\)
\( = \int {\ln } x \times \frac{1}{{{{\log }_e}b}}dx\)
\( = \frac{1}{{{{\log }_e}b}}\int {\ln } xdx\)
\( = \frac{1}{{{{\log }_e}b}}\left[ {\ln x \times x – \int {\frac{1}{x}} \times xdx} \right] + C\)
\( = \frac{1}{{{{\log }_e}b}}\left[ {\ln x \times x – \int d x} \right] + C\)
\( = \frac{1}{{{{\log }_e}b}}[x\ln x – x] + C\)
\( = x\left[ {\frac{1}{{{{\log }_e}b}} \times \ln x – \frac{1}{{{{\log }_e}b}}} \right] + C\)
\( = x\left[ {{{\log }_b}e \times {{\log }_e}x – \frac{1}{{{{\log }_e}b}}} \right] + C\)
\( = x\left[ {{{\log }_b}x – \frac{1}{{{{\log }_e}b}}} \right] + C\)
Example: \(\int {{{\log }{10}}} (x)dx = x\left( {{{\log }{10}}(x) – \frac{1}{{\ln (b)}}} \right) + c\) 

Some Other Properties

Few more properties of the logarithm functions are given below:

1. \({\log _{\rm{b}}}b = 1\) 
2. \({\log _b}1 = 0\) 
3. \({\log _b}0 = \) undefined
4. The logarithm of a negative number is not defined

How to Use the Logarithm Properties?

(i) We use the Logarithm properties to make the calculations simpler.
(ii) Logarithms help us to convert the exponential form \({2^5} = 32\) into logarithmic form \({\log _2}32 = 5\).
(iii) To write the product as the sum, you can use the logarithm property.
(iv) \(\log 14 = \log (7 \times 2) = \log 7 + \log 2\)
(v) The logarithm property is used to mention the division as the difference.
(vi) \(\log 0.3 = \log \frac{3}{{10}} = \log 3 – \log 10\)

The logarithm property is used to mention the exponent as the product.

\(\log \sqrt 5 = \log {5^{\frac{1}{2}}} = \frac{1}{2}\log 5\)

The logarithm property is used to split the bigger number into the smaller factors.

\(\log 24 = \log (8 \times 3) = \log \left( {{2^3} \times 3} \right) = \log {2^3} + \log 3 = 3\log 2 + \log 3\)

Properties of Logarithms Calculator

A logarithm calculator is an online math calculator that helps us to calculate the log value for the positive real number concerning the given or natural base logarithmic values. Using this calculator, we will understand how to find the logarithm of any number concerning the given base.

It is essential to follow the following steps:

1. First, you have to enter the number and the base of the logarithm. The values you are writing must be positive real numbers or the parameters.
2. Now, press the button CALCULATE to make the computation.
3. Here, the logarithm calculator will give the output logarithm of the positive real number

Input: Two positive real numbers or parameters. The first number is which the log will be calculated, and the second number is the number that is the base of the log.
Output: A positive real number.

Solved Examples

Q.1. Evaluate the given expression: \({\log _2}8 + {\log _2}4\).
Ans: Here, we will apply the rule,
\({\log _2}8 + {\log _2}4 = {\log _2}(8 \times 4)\)
\( = {\log _2}32\)
Now, rewrite \(32\) in the exponential form to get the value of its exponent.
\(32 = {2^5}\)
So, we have, \({\log _2}32 = {\log _2}{2^5} = 5{\log _2}2 = 5 \times 1 = 5\)
Hence, the required answer is \(5\).

Q.2. Evaluate \({\log _3}162 – {\log _3}2\).
Ans: Here, you have to apply the quotient rule law.
\({\log _3}162 – {\log _3}2 = {\log _3}\left( {\frac{{162}}{2}} \right) = {\log _3}81\)
Now, write \(81\) in the exponential form,
\(81 = {3^4}\)
So, we have, \({\log _3}81 = {\log _3}{3^4} = 4{\log _3}3 = 4 \times 1 = 4\)
Hence, the required answer is \(4\).

Q.3. Find the value of \(x\) where, \(2\log x = 4\log 3\)
Ans: Given, \(2\log x = 4\log 3\)
Here, you have to divide each side by the number \(2\).
\( \Rightarrow \log x = \frac{{4\log 3}}{2}\)
\( \Rightarrow \log x = 2\log 3\)
\( \Rightarrow \log x = \log {3^2}\)
\( \Rightarrow \log x = \log 9\)
Hence, \(x = 9\).

Q.4. Calculate \(\log x + \log (x – 1) = \log (3x + 12)\) 
Ans: Given, \(\log x + \log (x – 1) = \log (3x + 12)\)
\( \Rightarrow \log [x(x – 1)] = \log (3x + 12)\)
Now, you have to drop the logarithms to get:
\( \Rightarrow [x(x – 1)] = (3x + 12)\)
Here, you have to use the distributive property to remove the brackets.
\( \Rightarrow {x^2} – x = 3x + 12\)
\( \Rightarrow {x^2} – x – 3x – 12 = 0\)
\( \Rightarrow {x^2} – 4x – 12 = 0\)
\( \Rightarrow (x – 6)(x + 2) = 0\)
\( \Rightarrow x = – 2,\,x = 6\)
The logarithm is not defined for negative real numbers. Hence, the required answer is \(x = 6\).

Q.5. Find \(x\): \({\log _2}(5x + 6) = 5\)
Ans: Given,  \({\log _2}(5x + 6) = 5\)
Now, rewrite the given equation in the exponential form,
\({2^5} = 5x + 6\)
Simplify, \(32 = 5x + 6\)
Here, subtract \(6\) from both sides of the equation.
\(32 – 6 = 5x + 6 – 6\)
\( \Rightarrow 26 = 5x\) Find
\( \Rightarrow x = \frac{{26}}{5}\)
Hence, the required value of \(x\) is \(\frac{{26}}{5}\).

Q.6. Express \({5^3} = 125\) in logarithm form.
Ans: We have, \({5^3} = 125\)
Now we need to write in the logarithm form
We know that,
\({a^b} = c \Rightarrow {\log _a}c = b\)
So, we will write as \({\log _5}125 = 3\)
Hence, the required answer is given above.

Q.7. Express \({\log _{10}}1 = 0\) in exponential form.
Ans: We have, \({\log _{10}}1 = 0\)
Now we need to write in the exponential form
By the rule, we know that,
\({\log _a}c = b \Rightarrow {a^b} = c\)
So, we will write as \({10^0} = 1\)
Hence, the required answer is given above.

Q.8. Solve for \(x\) if \(\log (x – 1) + \log (x + 1) = {\log _2}1\)
Ans: We have \(\log (x – 1) + \log (x + 1) = {\log _2}1\)
\(\log (x – 1) + \log (x + 1) = 0\)
\(\log [(x – 1)(x + 1)] = 0\)
Now, we know that \(\log 1 = 0\)
\((x – 1)(x + 1) = 1\)
This becomes \({x^2} – 1 = 1\)
\({x^2} = 2\)
\(x = \pm \sqrt 2 \)
Thus, we know that log of the negative number is not defined.
Hence, the required answer is \(x = \sqrt 2 \).

Q.9. Express \(\log \left( {\frac{{75}}{{16}}} \right) – 2\log \left( {\frac{5}{9}} \right) + \log \left( {\frac{{32}}{{243}}} \right)\) in terms of \(2\) and \(3\).
Ans: We have  \(\log \left( {\frac{{75}}{{16}}} \right) – 2\log \left( {\frac{5}{9}} \right) + \log \left( {\frac{{32}}{{243}}} \right)\)
We know that \(tn{\log _a}m = {\log _a}{m^n}\)
\( \Rightarrow \log \left( {\frac{{75}}{{16}}} \right) – \log {\left( {\frac{5}{9}} \right)^2} + \log \left( {\frac{{32}}{{243}}} \right)\)
\( \Rightarrow \log \left( {\frac{{75}}{{16}}} \right) – \log \left( {\frac{{25}}{{81}}} \right) + \log \left( {\frac{{32}}{{243}}} \right)\)
Now, we know \({\log _a}m – {\log _a}n = {\log _a}\left( {\frac{m}{n}} \right)\)
\( \Rightarrow \log \left[ {\left( {\frac{{75}}{{16}}} \right) \div \left( {\frac{{25}}{{81}}} \right)} \right] + \log \left( {\frac{{32}}{{243}}} \right)\)
\( \Rightarrow \log \left[ {\frac{{75}}{{16}} \times \frac{{81}}{{25}}} \right] + \log \left( {\frac{{32}}{{243}}} \right)\)
\( \Rightarrow \log \left( {\frac{{243}}{{16}}} \right) + \log \left( {\frac{{32}}{{243}}} \right)\)
Thus, \({\log _a}m + {\log _a}n = {\log _a}mn\)
\( \Rightarrow \log \left( {\frac{{32}}{{16}}} \right)\)
\( \Rightarrow \log 2\)
Hence, the required answer is given above.

Summary

In the given article, we discussed logarithms with examples and defined logarithms. We also explained the rules and the properties of the logarithms with an example for each rule and property. We have also talked about a few points on how to use the logarithm properties. You can also find the solved examples and a few FAQs.

FAQs

Q.1. What are the four properties of a logarithm?
Ans: The four basic properties of the logarithm are given below:
(i) Product rule
(ii) Division rule
(iii) Power rule or the exponential rule
(iv) Change of base rule

Q.2. What are the properties of logarithms?
Ans: The properties of the logarithm are given below:
(i) Product rule: \(\log x + \log y = \log (x \times y) = \log xy\) 
(ii) Division rule: \({\log _b}\left( {\frac{m}{n}} \right) = {\log _b}m – {\log _b}n\) 
(iii) Power rule or the exponential rule: \({\log {\rm{b}}}\left( {{m^n}} \right) = n{\log {\rm{b}}}m\) 
(iv) Change of base rule: \({\log b}m = \frac{{{{\log }{\rm{a}}}m}}{{{{\log }_a}b}}\) 
(v) Base switch rule: \({\log _b}(a) = \frac{1}{{{{\log }_a}(b)}}\)

Q.3. What are the properties of logarithms and examples?
Ans: We know that in logarithm \({b^y} = a\) can be expressed as \({\log _b}a = y\).
Examples:
(i) \({2^{ – 3}} = \frac{1}{8} \Leftrightarrow {\log _2}\left( {\frac{1}{8}} \right) = – 3\) 
(ii) \({10^{ – 2}} = 0.01 \Leftrightarrow {\log _{10}}(0.01) = – 2\) 
(iii) \({2^6} = 64 \Leftrightarrow {\log _2}64 = 6\) 
(iv) \({3^2} = 9 \Leftrightarrow {\log _3}9 = 2\) 
(v) \({5^4} = 625 \Leftrightarrow {\log _5}625 = 4\) 
(vi) \({7^0} = 1 \Leftrightarrow {\log _7}1 = 0\) 
(vii) \({3^{ – 4}} = \frac{1}{{{3^4}}} = \frac{1}{{81}} \Leftrightarrow {\log _3}\frac{1}{{81}} = – 4\)

Q.4. What is the power property of log?
Ans: In this rule, the logarithm of a number \(m\) to the power (exponent) is equal to the exponent times its logarithm of the number \(m\).
Example: \({\log _b}\left( {{m^n}} \right) = n{\log _b}m\)

Q.5. What is the use of logarithms?
Ans: Logarithms are the most straightforward way to express large numbers. Mathematical calculations involving large numbers becomes easy if we use logarithm.

Q.6. What are the principles of the logarithm?
Ans: In exponential form, if we have \({b^x} = n\), then in the logarithmic form we can write \(x = {\log _b}n\). This is the principle of working of logarithm.
For example, \({2^3} = 8\);
Thus, \(3\) is the logarithm of \(8\) to base \(2\), or \(3 = {\log _2}8\)

Q.7. How many types of logarithms are there?
Ans: There are two types of logarithms, and they are as following:
(i) Common Logarithm: The common logarithm is called the base ten logarithms. It is written as \({\log _{10}}p\log p\). So, when the logarithm is taken involving base \(10\), we call it the common logarithm.
Example: \({10^2} = 100 \Rightarrow {\log _{10}}100 = 2\)
(ii) Natural Logarithm: The natural logarithm is known as the base \(e\) logarithm, where \(e\) is the Euler’s constant, which is approximately equal to \(2.71828\).
The natural logarithm is written as \(\ln x\) or \({\log _e}x\).
Example: \({2.71828^4} = 54.6 \Rightarrow {e^4} = 54.6\) or, \({\log _e}54.6 = 4\) or simply \(\ln 56.6 = 4\)

We hope you find this detailed article on the properties of logarithms helpful. If you have doubts or queries on this topic, feel to ask us in the comment section and we will be ready to help you at the earliest.