NZ Level 7 (NZC) Level 2 (NCEA)
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Logarithm Laws and Logarithm Properties
Lesson

Using addition and subtraction properties to simplify expressions

The addition property of logarithms relates the sum of two logarithms to the logarithm of a product.

Similarly, the subtraction property of logarithms relates the difference of two logarithms to the logarithm of a quotient.

Addition and subtraction property of logarithms

The addition property of logarithms is given by:

$\log_bx+\log_by=\log_b\left(xy\right)$logbx+logby=logb(xy)

The subtraction property of logarithms is given by:

$\log_bx-\log_by=\log_b\left(\frac{x}{y}\right)$logbxlogby=logb(xy)

We can prove these properties by using the corresponding properties of exponentials:

Proof of addition property:

We start by letting $\log_bx=N$logbx=N and $\log_by=M$logby=M. We can rewrite these two equations in their equivalent exponential forms, $b^N=x$bN=x and $b^M=y$bM=y. Multiplying these two expressions gives us the result:

$xy$xy $=$= $b^N\times b^M$bN×bM (Writing down the product)
  $=$= $b^{N+M}$bN+M (Using a property of exponentials)
$\log_b\left(xy\right)$logb(xy) $=$= $N+M$N+M (Rewriting in logarithmic form)
$\log_b\left(xy\right)$logb(xy) $=$= $\log_bx+\log_by$logbx+logby (Substituting)

Proof of subtraction property:

We follow a similar procedure and start by letting $\log_bx=N$logbx=N and $\log_by=M$logby=M. We can rewrite these in their exponential forms, $b^N=x$bN=x and $b^M=y$bM=y. Taking the quotient of the two expressions gives us the result:

$\frac{x}{y}$xy $=$= $\frac{b^N}{b^M}$bNbM (Writing down the quotient)
  $=$= $b^{N-M}$bNM (Using a property of exponentials)
$\log_b\left(\frac{x}{y}\right)$logb(xy) $=$= $N-M$NM (Rewriting in logarithmic form)
$\log_b\left(\frac{x}{y}\right)$logb(xy) $=$= $\log_bx-\log_by$logbxlogby (Substituting)

These two properties are especially valuable if we want to simplify expressions or solve equations involving logarithms.

 

Worked examples

example 1

Simplify the logarithmic expression $\log_310-\log_32$log310log32.

Think: Since the two logarithms have the same base, we can use the subtraction property of logarithms.

Do: To use the subtraction property of two logarithms, we can divide the arguments:

$\log_310-\log_32$log310log32 $=$= $\log_3\left(\frac{10}{2}\right)$log3(102) (Using the subtraction property)
  $=$= $\log_35$log35 (Simplifying the argument)
example 2

Rewrite $\log_56x$log56x as the sum or difference of two logarithms.

Think: Since there is a product in the logarithm, we can use the addition property in reverse.

Do: So using the addition property, we can rewrite $\log_56x$log56x in the form:

$\log_56+\log_5x$log56+log5x

 

Practice questions

question 1

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 2

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 3

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

Using power property to simplify expressions

We've already seen how to simplify logarithms using the addition and subtraction property. 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) $=$= $\log_a\left(x\times x\right)$loga(x×x) Rewrite $x^2$x2 as a product, $x\times x$x×x.
  $=$= $\log_ax+\log_ax$logax+logax

Use the addition property of logarithms,

 $\log_a\left(xy\right)=\log_ax+\log_ay$loga(xy)=logax+logay.

  $=$= $2\log_ax$2logax

Collect logarithms with the same base and variables.

We can also simplify logarithms with powers using the power property of 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

Now, let's simplify $\log_a\left(x^2\right)$loga(x2) using the power property.

$\log_a\left(x^2\right)$loga(x2) $=$= $2\log_ax$2logax

Use the power property$\log_a\left(x^n\right)=n\log_ax$loga(xn)=nlogax.

Notice this gives the exact same result as using the addition property. 

Let's consider the proof of the power property of logarithms:

Proof          
Let $x$x $=$= $a^m$am  
$x^n$xn $=$= $\left(a^m\right)^n$(am)n Raise both sides of $x=a^m$x=am to the power $n$n.
$x^n$xn $=$= $a^{mn}$amn Use the index law $\left(a^m\right)^n=a^{mn}$(am)n=amn.
$\log_a\left(x^n\right)$loga(xn) $=$= $mn$mn Express as a logarithm.
$\log_a\left(x^n\right)$loga(xn) $=$= $n\log_ax$nlogax Substitute back for $m=\log_ax$m=logax.

 

Worked example

Simplify the expression $\log_2\left(x^b\right)$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 of 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) $=$= $b\log_2x$blog2x

 

Practice questions

QUESTION 4

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

Write your answer without any powers.

Question 5 

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

Write your answer without any powers.

QUESTION 6

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

Write your answer without any powers.

Use multiple laws to simplify expressions

So far we've seen some properties of logarithms in isolation, and looked at how they may be individually useful to simplify an expression or solve an equation.

However, sometimes simplifying an expression may require using several of these properties.

The properties we have looked at are summarised below.

Properties of logarithms

The addition property of logarithms is given by:

$\log_bx+\log_by=\log_b\left(xy\right)$logbx+logby=logb(xy)

The subtraction property of logarithms is given by:

$\log_bx-\log_by=\log_b\left(\frac{x}{y}\right)$logbxlogby=logb(xy)

The power property of logarithms is given by:

$\log_b\left(x^n\right)=n\log_bx$logb(xn)=nlogbx

Some useful identities of logarithms are:

$\log_b1=0$logb1=0 and $\log_bb=1$logbb=1

 

Worked examples

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 subtraction property.

Do: To use the subtraction property, we take the quotient of the two arguments as follows:

$\log_3\left(100x^3\right)-\log_3\left(4x\right)$log3(100x3)log3(4x) $=$= $\log_3\left(\frac{100x^3}{4x}\right)$log3(100x34x) (Using the subtraction property)
  $=$= $\log_3\left(25x^2\right)$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 property)

 

Practice questions

question 7

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

  1. $\log_{10}10+\frac{\log_{10}\left(15^{20}\right)}{\log_{10}\left(15^5\right)}$log1010+log10(1520)log10(155)

  2. $\frac{8\log_{10}\left(\sqrt{10}\right)}{\log_{10}\left(100\right)}$8log10(10)log10(100)

question 8

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

 

Outcomes

M7-6

Manipulate rational, exponential, and logarithmic algebraic expressions

91261

Apply algebraic methods in solving problems

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