# 6.05 Extension: Properties of logarithms

Lesson

## Laws of exponents and laws of logarithms

Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.

$\log_b1=0$logb1=0

$\log_bb=1$logbb=1

Why are the above true? To find out, change the logarithm to an exponential: $\log_b1=0$logb1=0 becomes $b^0=1$b0=1  and $\log_bb=1$logbb=1 becomes $b^b=1$bb=1

Now, lets review the inverse property:

$\log_bb^x=x$logbbx=x

$b^{\log_bx}=x$blogbx=x$x>0$x>0

Why is this true? Convert each into its opposite: $\log_bb^x=x$logbbx=x becomes $b^x=b^x$bx=bx and $b^{\log_bx}=x$blogbx=x becomes $\log_bx=\log_bx$logbx=logbx

One-to-one

$\log_bM=\log_bN$logbM=logbN if and only if $M=N$M=N

This property is used to solve equations.

Remember, all properties associated with logarithms can be applied to natural logarithms.

#### Worked example

##### Question 1

What is the value of $x$x for the equation $\log_33x=\log_3\left(2x+5\right)$log33x=log3(2x+5)?

Think: Does the logarithms have the same base? Remember, we can not work with logarithms with unlike bases. They do have the same base. So, that means, just as with exponential equations with the same bases, we can set the arguments equal to each other.

Do:

 $\log_33x$log3​3x $=$= $\log_3\left(2x+5\right)$log3​(2x+5) (Given) $3x$3x $=$= $2x+5$2x+5 (Use one-to-one property to set arguments equal) $x$x $=$= $5$5 (Subtract both sides by $2x$2x)

### Product rule for logarithms

Recall the product rule for exponents:

$a^na^m=a^{n+m}$anam=an+m

The product rule for logarithms is similar:

$\log_bMN=\log_bM+\log_bN$logbMN=logbM+logbN

Why is this true? Lets use the rules we already know to prove this:

 $\log_bMN$logb​MN $=$= $\log_bM+\log_bN$logb​M+logb​N (Given) $b^{\log_bM+\log_bN}$blogb​M+logb​N $=$= $MN$MN (Rewrite as an exponential) $b^{\log_bM}b^{\log_bN}$blogb​Mblogb​N $=$= $MN$MN (Use laws of exponent: product rule) $MN$MN $=$= $MN$MN (Use inverse property for logarithms)

The product rule for logarithms can be applied repeatedly. The expression $\log_btuv$logbtuv can be rewritten as $\log_bt+\log_bu+\log_bv$logbt+logbu+logbv.

#### Worked example

##### question 2

Rewrite $\log_56x$log56x as two logarithms.

Think: Since there is a product in the logarithm, we can use the product rule for logarithms.

Do: So using the product rule for logarithms, we can rewrite $\log_56x$log56x in the form:

$\log_56+\log_5x$log56+log5x

### Quotient rule for logarithms

Recall the quotient rule for exponents:



The quotient rule for logarithms is similar:

$\log_b\left(\frac{M}{N}\right)=\log_bM-\log_bN$logb(MN)=logbMlogbN

Why is this true? Lets take a look using properties we already know:

 $\log_b\left(\frac{M}{N}\right)$logb​(MN​) $=$= $\log_bM-\log_bN$logb​M−logb​N (Given) $b^{\log_bM-\log_bN}$blogb​M−logb​N $=$= $\frac{M}{N}$MN​ (Rewrite as an exponential) $\frac{b^{\log_bM}}{b^{\log_bN}}$blogb​Mblogb​N​ $=$= $\frac{M}{N}$MN​ (Use laws of exponent: quotient rule) $\frac{M}{N}$MN​ $=$= $\frac{M}{N}$MN​ (Use inverse property for logarithms)

#### Worked example

##### question 3

Rewrite the logarithmic expression $\log_310-\log_32$log310log32 as one logarithmic expression.

Think: Since the two logarithms have the same base and they are subtracting, we can use the quotient rule for logarithms.

Do: To use the quotient rule for logarithms, we can divide the arguments:

 $\log_310-\log_32$log3​10−log3​2 (Given) $\log_3\left(\frac{10}{2}\right)$log3​(102​) (Use quotient rule for logarithms) $\log_35$log3​5 (Simplify the argument)

#### Practice questions

##### question 4

Simplify each of the following expressions without using a calculator. Leave answers in exact form.

1. $\log_{10}11+\log_{10}2+\log_{10}9$log1011+log102+log109

2. $\log_{10}12-\left(\log_{10}2+\log_{10}3\right)$log1012(log102+log103)

##### question 5

Simplify each of the following expressions without using a calculator. Leave answers in exact form.

1. $\log_{10}18-\log_{10}3$log1018log103

2. $\log_{10}7-\log_{10}28$log107log1028

##### question 6

Express $\log\left(\frac{pq}{r}\right)$log(pqr) as the sum and difference of log terms.

### Power rule for logarithms

We have already seen how to simplify logarithms using the product and quotient properties. Through the definition of logarithms we know that $x=a^m$x=am and $m=\log_ax$m=logax are equivalent. We are able to use this definition to discover some more helpful properties of logarithms such as the power property.

#### Exploration

Let's simplify $\log_a\left(x^2\right)$loga(x2) using the logarithmic properties that we already know.

 $\log_a\left(x^2\right)$loga​(x2) (Given) $\log_a\left(x\times x\right)$loga​(x×x) (Rewrite $x^2$x2as a product, $x\times x$x×x) $\log_ax+\log_ax$loga​x+loga​x (Use the product rule for logarithms, $\log_a\left(xy\right)=\log_ax+\log_ay$loga​(xy)=loga​x+loga​y $2\log_ax$2loga​x (Collect logarithms with the same base and variables.)

We can also simplify logarithms with powers using the power rule for logarithms, this property can be used for any values of the power $n$n.

Power property of logarithms

$\log_a\left(x^n\right)=n\log_ax$loga(xn)=nlogax

#### Worked example

##### question 7

Rewrite $\log_2\left(x^b\right)$log2(xb) using properties of logarithms. Write your answer without any powers.

Think: The subject of the logarithm has a power, this means we can use the power rule for logarithms,$\log_a\left(x^n\right)=n\log_ax$loga(xn)=nlogax.

What do the values of $a$a and $n$n represent?

Do: In this case $a=2$a=2 and $n=b$n=b. So we can bring the power down to the front, and then multiply it with the logarithm.

 $\log_2\left(x^b\right)$log2​(xb) $=$= $b\log_2x$blog2​x

#### Practice questions

##### QUESTION 8

Use the properties of logarithms to rewrite the expression $\log_4\left(x^7\right)$log4(x7).

##### Question 9

Use the properties of logarithms to rewrite the expression $\log\left(\left(x+6\right)^5\right)$log((x+6)5).

##### QUESTION 10

Use the properties of logarithms to rewrite $\log\left(\left(3x\right)^5\right)$log((3x)5) as the sum of two logarithms.

### Condensing and distributing logarithmic expressions

When viewed together, the product rule, quotient rule and power rule are often called "laws of logs." We can apply more than one rule to simplify an expression.

#### Worked examples

##### question 11

Simplify the expression $\log_3\left(100x^3\right)-\log_3\left(4x\right)$log3(100x3)log3(4x), writing your answer as a single logarithm.

Think: Each logarithm in the expression has the same base, so we can express the difference as a single logarithm using the quotient rule.

Do: To use the quotient rule, we divide the two arguments as follows:

 $\log_3\left(100x^3\right)-\log_3\left(4x\right)$log3​(100x3)−log3​(4x) (Given) $\log_3\left(\frac{100x^3}{4x}\right)$log3​(100x34x​) (Use the quotient rule) $\log_3\left(25x^2\right)$log3​(25x2) (Simplifying the argument) $\log_3\left(\left(5x\right)^2\right)$log3​((5x)2) (Rewriting the argument as a power) $2\log_3\left(5x\right)$2log3​(5x) (Using the power rule)

##### question 12

Distribute the expression: $\ln\frac{x^2y^3}{z^4}$lnx2y3z4

Think: List the operations within the argument: exponents, multiplication, division. We are going to start distributing with division because of the sensitivity to order.

Do:

 $\ln\frac{x^2y^3}{z^4}$lnx2y3z4​ (Given) $\ln x^2y^3-\ln z^4$lnx2y3−lnz4 (Use the quotient rule) $\ln x^2+\ln y^3-\ln z^4$lnx2+lny3−lnz4 (Use the product rule) $2\ln x+3\ln y-4\ln z$2lnx+3lny−4lnz (Use power rule)

#### Practice questions

##### Question 13

Express $3\ln\left(x^5\right)-4\ln\left(x^2\right)$3ln(x5)4ln(x2) as a single log expression.

##### question 14

Express $5\log x+3\log y$5logx+3logy as a single logarithm.

## Changing the base

We often encounter occasions where we need to take a logarithm given in one base and express it as a logarithm in another base. A change of base formula has been developed to do just that.

Change of Base Rule!

$\log_ab=\frac{\log_cb}{\log_ca}=\frac{1}{\log_ba}$logab=logcblogca=1logba

#### Worked examples

##### question 15

Change the base of the logarithm to $2$2: $\log_48$log48 and evaluate.

Think: Use the change of base rule.

Do:

 $\log_48$log4​8 (Given) $\frac{\log_28}{\log_24}$log2​8log2​4​ (Use the change of base rule) $\frac{\log_22^3}{\log_22^2}$log2​23log2​22​ (Rewrite the argument as exponents) $\frac{3}{2}$32​ (Use the inverse property)

##### question 16

Evaluate $\log_9564$log9564 using by changing the base to $10$10 . Round to 3 decimal points.

Think: Review the change of base rule. Can we rewrite the current base or argument as exponents with base $10$10?

Do:

 $\log_9564$log9​564 (Given) $\frac{\log564}{\log9}$log564log9​ (Use change of base rule) $2.883$2.883 (Cannot write the arguments as exponents with base $10$10, use calculator to find $\log564$log564 and $\log9$log9 then divide)

#### Practice questions

##### QUESTION 17

Rewrite $\log_416$log416 in terms of base $10$10 logarithms.

##### QUESTION 18

Rewrite $\log_320$log320 in terms of base $4$4 logarithms.

##### QUESTION 19

Rewrite $\log_3\sqrt{5}$log35 in terms of base $10$10 logarithms.