# Intro to logarithm properties

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

(These properties apply for any values of $M$, $N$, and $b$ for which each logarithm is defined, which is $M$, $N>0$ and $0.)
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 $1$.

#### 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: $\log_b(MN)=\log_b(M)+\log_b(N)$

This property says that the logarithm of a product is the sum of the logs of its factors.
Sure! If $M=4$, $N=8$ and $b=2$, then according to the property, $\log_2(4\cdot 8)=\log_2(4)+\log_2(8)$.
The work below shows that the property is indeed true in this case!
\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 $\log_6(5y)$.
Notice that the two factors of the argument of the logarithm are $\blueD 5$ and $\greenD y$. We can directly apply the product rule to expand the log.
\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 $\log_3(10)+\log_3(x)$.
Since the two logarithms have the same base (base-$3$), we can apply the product rule in the reverse direction:
\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 $\log_2(8)+\log_3(y)$.

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

To expand $\log_2(3a)$ means to write the expression as the sum of two logarithms.
We can directly apply the product rule to expand the log.
\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 $\log_5(2y)+\log_5(8)$.

To condense $\log_5(2y)+\log_5(8)$ means to write the expression as a single logarithm. Since the bases are the same, we can apply the product rule.
\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: $\log_b\left(\dfrac{M}{N}\right)=\log_b(M)-\log_b(N)$

This property says that the log of a quotient is the difference of the logs of the dividend and the divisor.
Sure! If $M=81$, $N=3$ and $b=3$, then according to the property, $\log_3\left(\dfrac{81}{3}\right)=\log_3(81)-\log_3(3)$.
The work below shows that the property is indeed true in this case!
\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 $\log_7\left(\dfrac{a}{2}\right)$, writing it as the difference of two logarithms by directly applying the 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 $\log_4(x^3)-\log_4(y)$.
Since the two logarithms have the same base (base-$4$), we can apply the quotient rule in the reverse direction:
\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 $\log_2(8)-\log_3(y)$.

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

We can write $\log_b\left(\dfrac{4}{c}\right)$ as the difference of two logarithms by directly applying the 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 $\log(3z)-\log(8)$.

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

# The power rule: $\log_b(M^p)=p\log_b(M)$

This property says that the log of a power is the exponent times the logarithm of the base of the power.
Sure! If $M=4$, $p=2$, and $b=4$, then according to the property, $\log_4\left(4^2\right)=2\log_4(4)$.
The work below shows that the property is indeed true in this case!
\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 $\log_2\left(x^3\right)$.
\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 $4\log_5(2)$,
When we condense a logarithmic expression using the power rule, we make any multipliers into powers.
\begin{aligned}\maroonC4\log_5(2)&=\log_5\left(2^\maroonC 4\right)~~&&\small{\gray{\text{Power rule}}}\\ \\ &=\log_5(16)\\ \end{aligned}

5) Expand $\log_7(x^5)$.

\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 $6\ln(y)$.

\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 $\log_b\left(\dfrac{2x^3}{5}\right)$?

First, apply the 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.
\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.
\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 $\log_b(2)+3\log_b(x)-\log_b(5)$.
2) Which of the following is equivalent to $3\log_2(x)-2\log_2(5)$?
\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}
\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 $\log_2\left(\dfrac{{x^3}}{{25}}\right)$