Learn about the properties of logarithms and how to use them to rewrite logarithmic expressions. For example, expand log₂(3a).
The product rulelogb(MN)=logb(M)+logb(N)\large\log_b(MN)=\log_b(M)+\log_b(N)
The quotient rulelogb(MN)=logb(M)logb(N)\large\log_b\left(\frac{M}{N}\right)=\log_b(M)-\log_b(N)
The power rulelogb(Mp)=plogb(M)\large\log_b(M^p)=p\log_b(M)

(These properties apply for any values of MM, NN, and bb for which each logarithm is defined, which is MM, N>0N>0 and 0<b10<b\neq1.)
Remember, in order for a logarithm to be defined, the argument of the logarithm must be positive and the base of the logarithm must also be positive and not equal to 11.

What you should be familiar with before taking this lesson

You should know what logarithms are. If you don't, please check out our intro to logarithms.

What you will learn in this lesson

Logarithms, like exponents, have many helpful properties that can be used to simplify logarithmic expressions and solve logarithmic equations. This article explores three of those properties.
Let's take a look at each property individually.

The product rule: logb(MN)=logb(M)+logb(N)\log_b(MN)=\log_b(M)+\log_b(N)

This property says that the logarithm of a product is the sum of the logs of its factors.
Sure! If M=4M=4, N=8N=8 and b=2b=2, then according to the property, log2(48)=log2(4)+log2(8)\log_2(4\cdot 8)=\log_2(4)+\log_2(8).
The work below shows that the property is indeed true in this case!
log2(48)=log2(4)+log2(8)log2(32)=log2(4)+log2(8)Since 48=325=2+3Evaluating the logs5=5\begin{aligned}\log_2({4\cdot 8})&=\log_2(4)+\log_2(8)\\ \\ \log_2(32)&=\log_2(4)+\log_2(8)&&\small{\gray{\text{Since $4\cdot8=32$}}}\\ \\ 5&=2+3&&\small{\gray{\text{Evaluating the logs}}}\\ \\ 5&=5\end{aligned}
This is by no means a proof! Rather, it may convince us that the property is plausible and perhaps give us some insight as to why this is true.
We can use the product rule to rewrite logarithmic expressions.

Example 1: Expanding logarithms

For our purposes, expanding a logarithm means writing it as the sum of two logarithms or more.
Let's expand log6(5y)\log_6(5y).
Notice that the two factors of the argument of the logarithm are 5\blueD 5 and y\greenD y. We can directly apply the product rule to expand the log.
log6(5y)=log6(5y)=log6(5)+log6(y)        Product rule\begin{aligned}\log_6(\blueD5\greenD y)&=\log_6(\blueD5\cdot \greenD y)\\ \\ &=\log_6(\blueD5)+\log_6(\greenD y)&&~~~~~~~~\small{\gray{\text{Product rule}}} \end{aligned}

Example 2: Condensing logarithms

For our purposes, compressing a sum of two or more logarithms means writing it as a single logarithm.
Let's condense log3(10)+log3(x)\log_3(10)+\log_3(x).
Since the two logarithms have the same base (base-33), we can apply the product rule in the reverse direction:
log3(10)+log3(x)=log3(10x)Product rule=log3(10x)\begin{aligned}\log_3(\blueD{10})+\log_3(\greenD x)&=\log_3(\blueD{10}\cdot \greenD x)&&\small{\gray{\text{Product rule}}}\\ \\ &=\log_3({10} x) \end{aligned}

An important note

When we compress logarithmic expressions using the product rule, the bases of all the logarithms in the expression must be the same.
For example, we cannot use the product rule to simplify something like log2(8)+log3(y)\log_2(8)+\log_3(y).

Check your understanding

1) Expand log2(3a)\log_2(3a).

To expand log2(3a)\log_2(3a) means to write the expression as the sum of two logarithms.
We can directly apply the product rule to expand the log.
log2(3a)=log2(3a)=log2(3)+log2(a)Product rule\begin{aligned}\log_2(\blueD3\greenD a)&=\log_2(\blueD3\cdot \greenD a)\\ \\ &=\log_2(\blueD3)+\log_2(\greenD a) &&\small{\gray{\text{Product rule}}} \end{aligned}
2) Condense log5(2y)+log5(8)\log_5(2y)+\log_5(8).

To condense log5(2y)+log5(8)\log_5(2y)+\log_5(8) means to write the expression as a single logarithm. Since the bases are the same, we can apply the product rule.
log5(2y)+log5(8)=log5(2y8)Product rule=log5(16y)\begin{aligned}\log_5(\blueD{2y})+\log_5(\greenD {8})&=\log_5(\blueD{2y}\cdot \greenD {8})&&\small{\gray{\text{Product rule}}}\\ \\ &=\log_5(16y) \end{aligned}

The quotient rule: logb(MN)=logb(M)logb(N) \log_b\left(\dfrac{M}{N}\right)=\log_b(M)-\log_b(N)

This property says that the log of a quotient is the difference of the logs of the dividend and the divisor.
Sure! If M=81M=81, N=3N=3 and b=3b=3, then according to the property, log3(813)=log3(81)log3(3)\log_3\left(\dfrac{81}{3}\right)=\log_3(81)-\log_3(3).
The work below shows that the property is indeed true in this case!
log3(813)=log3(81)log3(3)log3(27)=log3(81)log3(3)Since 81÷3=273=41Evaluating the logs3=3\begin{aligned} \\\log_3\left({\dfrac{81}{3}}\right)&=\log_3(81)-\log_3(3)\\ \\ \log_3(27)&=\log_3(81)-\log_3(3)&&\small{\gray{\text{Since $81\div 3=27$}}}\\ \\ 3&=4-1&&\small{\gray{\text{Evaluating the logs}}}\\ \\ 3&=3\end{aligned}
This is by no means a proof! Rather, it may convince us that the property is plausible and perhaps give us some insight as to why this is true.
Now let's use the quotient rule to rewrite logarithmic expressions.

Example 1: Expanding logarithms

Let's expand log7(a2)\log_7\left(\dfrac{a}{2}\right), writing it as the difference of two logarithms by directly applying the quotient rule.
log7(a2)=log7(a)log7(2)Quotient rule\begin{aligned}\log_7\left(\dfrac{\purpleC a}{\goldD 2}\right)&=\log_7(\purpleC a)-\log_7(\goldD 2) &\small{\gray{\text{Quotient rule}}} \end{aligned}

Example 2: Condensing logarithms

Let's condense log4(x3)log4(y)\log_4(x^3)-\log_4(y).
Since the two logarithms have the same base (base-44), we can apply the quotient rule in the reverse direction:
log4(x3)log4(y)=log4(x3y)Quotient rule\begin{aligned}\log_4(\purpleC{x^3})-\log_4(\goldD{y})&=\log_4\left(\dfrac{\purpleC{x^3}}{\goldD{y}}\right)&&\small{\gray{\text{Quotient rule}}}\\ \\ \end{aligned}

An important note

When we compress logarithmic expressions using the quotient rule, the bases of all logarithms in the expression must be the same.
For example, we cannot use the quotient rule to simplify something like log2(8)log3(y)\log_2(8)-\log_3(y).

Check your understanding

3) Expand logb(4c)\log_b\left(\dfrac{4}{c}\right).

We can write logb(4c)\log_b\left(\dfrac{4}{c}\right) as the difference of two logarithms by directly applying the quotient rule.
logb(4c)=logb(4)logb(c)Quotient rule\begin{aligned}\log_b\left(\dfrac{\purpleC 4}{\goldD c}\right)&=\log_b(\purpleC 4)-\log_b(\goldD c) &&\small{\gray{\text{Quotient rule}}} \end{aligned}
4) Condense log(3z)log(8)\log(3z)-\log(8).

Notice that there is no base indicated on the logarithms. This means that the base is 1010 since log(x)=log10(x)\log(x)=\log_{10}(x).
Since the base of each logarithm is the same, we can apply the quotient rule.
log(3z)log(8)=log(3z8)\begin{aligned}\log(\purpleC{3z})-\log(\goldD{8})&=\log\left(\dfrac{\purpleC{3z}}{\goldD{8}}\right)\\ \end{aligned}
Of course, log10(3z8)\log_{10}\left(\dfrac{3z}{8}\right) is also an acceptable answer.

The power rule: logb(Mp)=plogb(M)\log_b(M^p)=p\log_b(M)

This property says that the log of a power is the exponent times the logarithm of the base of the power.
Sure! If M=4M=4, p=2p=2, and b=4b=4, then according to the property, log4(42)=2log4(4)\log_4\left(4^2\right)=2\log_4(4).
The work below shows that the property is indeed true in this case!
log4(42)=2log4(4)log4(16)=2log4(4)Since 42=162=21Evaluating the logs2=2 \begin{aligned} \\\log_4\left({4^2}\right)&=2\log_4(4)\\ \\ \log_4(16)&=2\log_4(4)&&\small{\gray{\text{Since $4^2=16$}}}\\ \\ 2&=2\cdot 1&&\small{\gray{\text{Evaluating the logs}}}\\\\ \\ 2&=2 \end{aligned}
This is by no means a proof! Rather, it may convince us that the property is plausible and perhaps give us some insight as to why this is true.
Now let's use the power rule to rewrite log expressions.

Example 1: Expanding logarithms

For our purposes in this section, expanding a single logarithm means writing it as a multiple of another logarithm.
Let's use the power rule to expand log2(x3)\log_2\left(x^3\right).
log2(x3)=3log2(x)Power rule=3log2(x)\begin{aligned}\log_2\left(x^\maroonC3\right)&=\maroonC3\cdot \log_2(x)&&\small{\gray{\text{Power rule}}}\\ \\ &=3\log_2(x) \end{aligned}

Example 2: Condensing logarithms

For our purposes in this section, condensing a multiple of a logarithm means writing it as a another single logarithm.
Let's use the power rule to condense 4log5(2)4\log_5(2),
When we condense a logarithmic expression using the power rule, we make any multipliers into powers.
4log5(2)=log5(24)  Power rule=log5(16)\begin{aligned}\maroonC4\log_5(2)&=\log_5\left(2^\maroonC 4\right)~~&&\small{\gray{\text{Power rule}}}\\ \\ &=\log_5(16)\\ \end{aligned}

Check your understanding

5) Expand log7(x5)\log_7(x^5).

log7(x5)=5log7(x)Power rule=5log7(x)\begin{aligned}\log_7\left(x^\maroonC5\right)&=\maroonC5\cdot \log_7(x)&&\small{\gray{\text{Power rule}}}\\ \\ &=5\log_7(x) \end{aligned}
6) Condense 6ln(y)6\ln(y).

6ln(y)=ln(y6)Power rule\begin{aligned}\maroonC6\ln(y)&=\ln\left(y^\maroonC 6\right)&&\small{\gray{\text{Power rule}}}\\ \end{aligned}

Challenge problems

To solve these next problems, you will have to apply several properties in each case. Give it a try!
1) Which of the following is equivalent to logb(2x35)\log_b\left(\dfrac{2x^3}{5}\right)?
Choose 1 answer:
Choose 1 answer:

First, apply the quotient rule.
logb(2x35)=logb(2x3)logb(5)               Quotient rule\begin{aligned} \log_b\left(\dfrac{\blueD{2x^3}}{\greenD{5}}\right) &= \log_b(\blueD{2x^3})-\log_b(\greenD{5}) ~~~~~~~~~~~~~~~&&\small{\gray{\text{Quotient rule}}}\\\\ \end{aligned}
Next, notice that there is a product in the first term. Use the product rule to write this as the sum of two logarithms.
logb(2x35)=logb(2x3)logb(5)=logb(2)+logb(x3)logb(5) Product rule\begin{aligned}\phantom{\log_b\left(\dfrac{\blueD{2x^3}}{\greenD{5}}\right)} &=\log_b({\purpleC{2}\goldD {x^3}})-\log_b({5})\\\\ &= \log_b(\purpleC{2})+\log_b(\goldD {x^3})-\log_b(5)&&~\small{\gray{\text{Product rule}}} \\\\ \end{aligned}
Finally, notice that there is still a power in the expression. We can use the power rule to bring down this exponent.
logb(x3yz2)=logb(2)+logb(x3)logb(5)=logb(2)+3logb(x)logb(5)Power rule\begin{aligned}\phantom{\log_b\left(\dfrac{\blueD{x^3y}}{\greenD{z^2}}\right)} &=\log_b( 2)+\log_b({x^\maroonC 3})-\log_b(5)\\\\ &=\log_b(2)+\maroonC 3\log_b(x)- \log_b(5)&&\small{\gray{\text{Power rule}}}\\ \\ \end{aligned}
You know that a logarithmic expression is completely expanded when there are no powers, products or quotients remaining in the arguments of the logarithms.
So the expanded form is logb(2)+3logb(x)logb(5)\log_b(2)+3\log_b(x)-\log_b(5).
2) Which of the following is equivalent to 3log2(x)2log2(5)3\log_2(x)-2\log_2(5)?
Choose 1 answer:
Choose 1 answer:

To condense this logarithmic expression, let's first write the multiples of the logarithms as exponents using the power rule.
3log2(x)2log2(5)=log2(x3)log2(52)=log2(x3)log2(25)\begin{aligned}\maroonD 3\log_2(x)-\maroonD2\log_2(5)&=\log_2(x^\maroonD 3)-\log_2(5^\maroonD 2)\\ \\ &=\log_2(x^3)-\log_2(25)\\ \end{aligned}
Next use the quotient rule to write the expression as a single logarithm.
3log2(x2)2log2(5)=log2(x3)log2(25)=log2(x325)\begin{aligned}\phantom{\maroonD 3\log_2(x^2)-\maroonD2\log_2(5)}&=\log_2(\purpleC{x^3})-\log_2(\goldD{25})\\ \\ &=\log_2\left(\dfrac{\purpleC{x^3}}{\goldD{25}}\right)\\ \end{aligned}
The condensed form is log2(x325)\log_2\left(\dfrac{{x^3}}{{25}}\right)
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