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# 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${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$
The quotient rule${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)$
The power rule${\mathrm{log}}_{b}\left({M}^{p}\right)=p\cdot {\mathrm{log}}_{b}\left(M\right)$
(These properties apply for any values of $M$, $N$, and $b$ for which each logarithm is defined, which is $M$, $N>0$ and $0.)

#### 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: ${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$‍

This property says that the logarithm of a product is the sum of the logs of its factors.
We can use the product rule to rewrite logarithmic expressions.

### Example: Expanding logarithms using the product rule

For our purposes, expanding a logarithm means writing it as the sum of two logarithms or more.
Let's expand ${\mathrm{log}}_{6}\left(5y\right)$.
Notice that the two factors of the argument of the logarithm are $5$ and $y$. We can directly apply the product rule to expand the log.
$\begin{array}{rl}{\mathrm{log}}_{6}\left(5y\right)& ={\mathrm{log}}_{6}\left(5\cdot y\right)\\ \\ & ={\mathrm{log}}_{6}\left(5\right)+{\mathrm{log}}_{6}\left(y\right)& & \text{Product rule}\end{array}$

### Example: Condensing logarithms using the product rule

For our purposes, compressing a sum of two or more logarithms means writing it as a single logarithm.
Let's condense ${\mathrm{log}}_{3}\left(10\right)+{\mathrm{log}}_{3}\left(x\right)$.
Since the two logarithms have the same base (base-$3$), we can apply the product rule in the reverse direction:
$\begin{array}{rlrl}{\mathrm{log}}_{3}\left(10\right)+{\mathrm{log}}_{3}\left(x\right)& ={\mathrm{log}}_{3}\left(10\cdot x\right)& & \text{Product rule}\\ \\ & ={\mathrm{log}}_{3}\left(10x\right)\end{array}$

### 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 ${\mathrm{log}}_{2}\left(8\right)+{\mathrm{log}}_{3}\left(y\right)$.

### Check your understanding

1) Expand ${\mathrm{log}}_{2}\left(3a\right)$.

2) Condense ${\mathrm{log}}_{5}\left(2y\right)+{\mathrm{log}}_{5}\left(8\right)$.

## The quotient rule: ${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)$‍

This property says that the log of a quotient is the difference of the logs of the dividend and the divisor.
Now let's use the quotient rule to rewrite logarithmic expressions.

### Example: Expanding logarithms using the quotient rule

Let's expand ${\mathrm{log}}_{7}\left(\frac{a}{2}\right)$, writing it as the difference of two logarithms by directly applying the quotient rule.
$\begin{array}{rlr}{\mathrm{log}}_{7}\left(\frac{a}{2}\right)& ={\mathrm{log}}_{7}\left(a\right)-{\mathrm{log}}_{7}\left(2\right)& \text{Quotient rule}\end{array}$

### Example: Condensing logarithms using the quotient rule

Let's condense ${\mathrm{log}}_{4}\left({x}^{3}\right)-{\mathrm{log}}_{4}\left(y\right)$.
Since the two logarithms have the same base (base-$4$), we can apply the quotient rule in the reverse direction:
$\begin{array}{rlrl}{\mathrm{log}}_{4}\left({x}^{3}\right)-{\mathrm{log}}_{4}\left(y\right)& ={\mathrm{log}}_{4}\left(\frac{{x}^{3}}{y}\right)& & \text{Quotient rule}\end{array}$

### 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 ${\mathrm{log}}_{2}\left(8\right)-{\mathrm{log}}_{3}\left(y\right)$.

### Check your understanding

3) Expand ${\mathrm{log}}_{b}\left(\frac{4}{c}\right)$.

4) Condense $\mathrm{log}\left(3z\right)-\mathrm{log}\left(8\right)$.

## The power rule: ${\mathrm{log}}_{b}\left({M}^{p}\right)=p{\mathrm{log}}_{b}\left(M\right)$‍

This property says that the log of a power is the exponent times the logarithm of the base of the power.
Now let's use the power rule to rewrite log expressions.

### Example: Expanding logarithms using the power rule

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 ${\mathrm{log}}_{2}\left({x}^{3}\right)$.
$\begin{array}{rlrl}{\mathrm{log}}_{2}\left({x}^{3}\right)& =3\cdot {\mathrm{log}}_{2}\left(x\right)& & \text{Power rule}\\ \\ & =3{\mathrm{log}}_{2}\left(x\right)\end{array}$

### Example: Condensing logarithms using the power rule

For our purposes in this section, condensing a multiple of a logarithm means writing it as another single logarithm.
Let's use the power rule to condense $4{\mathrm{log}}_{5}\left(2\right)$,
When we condense a logarithmic expression using the power rule, we make any multipliers into powers.
$\begin{array}{rlrl}4{\mathrm{log}}_{5}\left(2\right)& ={\mathrm{log}}_{5}\left({2}^{4}\right)& & \text{Power rule}\\ \\ & ={\mathrm{log}}_{5}\left(16\right)\end{array}$

### Check your understanding

5) Expand ${\mathrm{log}}_{7}\left({x}^{5}\right)$.

6) Condense $6\mathrm{ln}\left(y\right)$.

## Challenge problems

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

8) Which of the following is equivalent to $3{\mathrm{log}}_{2}\left(x\right)-2{\mathrm{log}}_{2}\left(5\right)$?
Choose 1 answer:

## Want to join the conversation?

• Why isn't there a base in question 4?
(28 votes)
• Since both of them have no base you can presume they have the same base (no base) so you can apply the property of log. Your answer would be without base as well.

Later on you'll learn that no base log is understood as base of 10.
(118 votes)
• in the 1st challenging problem the ans given is (c) instead of a.if simplified
we get log base(b) (2x^3/5)= 3log base(b) 2+3log base(b) x - log base(b) 5. it shud be either a
(10 votes)
• Your mistake is in dealing with log_b (2x^3).
It is only the x that is cubed, not the 2.
So we first need to separate the factors (2 and x^3), getting
log_b (2) + log_b (x^3).
Only then can we deal with the cube, getting
log_b (2) + 3 log_b (x)
(47 votes)
• why the base cant be 1 of a logarithm?
(8 votes)
• Let take an example of log₁(25)=x
This would mean 1^x=25
But 1 to any powers will always be 1.
So the base can't be 1 because it would make the log expression false, unless log₁(1)=x but then x would be any and all real numbers. So the convention is to rule out log base 1.
(49 votes)
• Can the answer to logarithm be negative?
(3 votes)
• Any number less than 1 but greater than 0 will have a negative logarithm*
eg ln(0.1) = -2.303
(19 votes)
• why logarithms used in graphs ?
(6 votes)
• Interesting question as a fraction of people might get confused in it.

Log is the inverse function of exponent which you can say another form of exponent...so like we use exponents in graph eg ---> y = x^2 same we can also write log_x (y) = 2
And both are the same but at some point things get easier while using log.

A bit of a practice and experience and you will understand what I am saying.

Have a good day !!
(10 votes)
• What is the purpose of using log?
(2 votes)
• To find the unknown in an equation containing powers.
For example 2^b = 4. You can apply log2(4) = b
-> b = 2
(10 votes)
• why cant there be 1 for base
(4 votes)
• 1 raised to any power just equals 1.
Thus, the exponential equation: y = 1^x simplifies to y=1, a linear equation that creates a horizontal line.

If you apply this to logarithms, log_1(x), then the only possible value for "x" is 1 and log_1(1) has an infinite set of solutions.

Hope this helps.
(5 votes)
• Is there a distributive property for logarithms? For example, does ln(a+b)=ln(a)+ln(b)? I don't think I would use the product rule here. Is this covered in another lesson? Thanks!
(2 votes)
• Properties like this are covered later in the logarithms playlist. The property you suggest doesn't hold. ln(1+2)=ln(3), but ln(1)+ln(2)=0+ln(2)=ln(2), and ln(3)≠ln(2).

The relevant property here is that ln(ab)=ln(a)+ln(b). If you start from the property e^xe^y=e^(x+y), you can take the natural log of both sides to get

ln(e^xe^y)=x+y

Now let e^x=a (or x=ln(a)) and e^y=b (or y=ln(b)) and substitute in to get ln(ab)=ln(a)+ln(b).
(8 votes)
• What would X be in this equation: log4(x^3) + log4(x) = 5
(2 votes)
• log4(x^3) + log4(x) = 5
3 * log4(x) + log4(x) = 5
4 * log4(x) = 5
log4(x) = 5/4
x = 4^(5/4)
(7 votes)
• If 2log(x+4)-log(x+12)=log10, I get x=-8,4.
Do I have to eliminate -8 from the answer because the -8 makes (x+4) a negative number? Or I keep -8 because the coefficient attached to the log (x+4) turns (-4)^2=16?
(4 votes)
• You are correct that when x = -8, the expression (x + 4) becomes negative, which is not valid because the logarithm of a negative number is undefined. However, you should keep both answers (-8 and 4) and check them separately to see which one satisfies the original equation.

When x = 4, 2log(x+4) - log(x+12) = 2log(8) - log(16) = 2(3) - 1 = 5, and log(10) = 1, so the left-hand side of the equation matches the right-hand side. Therefore, x = 4 is a valid solution.

When x = -8, 2log(x+4) - log(x+12) = 2log(-4) - log(4) = undefined, so x = -8 is not a valid solution.

Therefore, the only solution to the equation is x = 4.
(3 votes)