20.06 Logarithmic laws (Extended)

Because of the link between indices and logarithms, it follows that these index laws can be manipulated to create a set of equivalent laws that apply to logarithms.

 

The addition and subtraction logarithm laws

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

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

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

Proof of addition law:

We start by letting $\log_bx=N$logb​x=N and $\log_by=M$logb​y=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)	$=$=	$\log_bx+\log_by$logb​x+logb​y	Substituting

 

Proof of subtraction law:

We follow a similar procedure and start by letting $\log_bx=N$logb​x=N and $\log_by=M$logb​y=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
	$=$=	$b^{N-M}$bN−M	Using a property of exponentials
$\log_b\left(\frac{x}{y}\right)$logb​(xy​)	$=$=	$N-M$N−M	Rewriting in logarithmic form
$\log_b\left(\frac{x}{y}\right)$logb​(xy​)	$=$=	$\log_bx-\log_by$logb​x−logb​y	Substituting

 

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

 

Worked examples

example 1

Simplify the logarithmic expression $\log_310-\log_32$log3​10−log3​2.

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

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

$\log_310-\log_32$log3​10−log3​2	$=$=	$\log_3\left(\frac{10}{2}\right)$log3​(102​)	Using the subtraction law
	$=$=	$\log_35$log3​5	Simplifying the argument

example 2

Rewrite $\log_56x$log5​6x as a sum or difference of two logarithms.

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

Do: So using the addition law, we can rewrite $\log_56x$log5​6x in the form:

$\log_56+\log_5x$log5​6+log5​x

Reflect: Note that we could have chosen any pair of factors that multiply to give $6x$6x to rewrite this expression. So, for example, another possible answer would be

$\log_52+\log_53x$log5​2+log5​3x

 

Practice questions

question 1

question 2

 

The power logarithm law

We've already seen how to simplify logarithms using the addition and subtraction property.We know that $x=a^m$x=am and $m=\log_ax$m=loga​x are equivalent. We are able to use this definition to discover some more helpful properties of logarithms such as the power law.

 

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.
	$=$=	$\log_ax+\log_ax$loga​x+loga​x	Use the addition law of 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 law of logarithms, this property can be used for any values of the power $n$n.

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

$\log_a\left(x^2\right)$loga​(x2)

$=$=

$2\log_ax$2loga​x

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

 

Let's consider the proof of the power law 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)			$=$=	$n\log_ax$nloga​x	Substitute back for $m=\log_ax$m=loga​x.

 

Practice questions

question 3

Question 4

Question 5

 

Using a combination of laws to simplify logarithmic expressions

So far we've seen some laws 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 laws.

The laws we have looked at are summarised below.

 

Worked example

Example 3

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 law.

Do: To use the subtraction law, 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 law
	$=$=	$\log_3\left(25x^2\right)$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 law

 

Practice questions

question 6

question 7

Question 8

 

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. For example, if we wanted to use the calculator to evaluate a logarithm but need to change the base to $10$10 or $e$e. A change of base formula has been developed to do just that.

Suppose we think of a number $y$y expressed as $y=\log_pa$y=logp​a.

Since $y=\log_pa$y=logp​a, then from the definition of a logarithm this means that:

$a$a

$=$=

$p^y$py

 

If we now take logarithms of some other base $q$q, we have that:

$\log_qa$logq​a	$=$=	$\log_q\left(p^y\right)$logq​(py)
$\log_qa$logq​a	$=$=	$y\log_qp$ylogq​p

From the working rules of logarithms this simplifies to:

$y$y

$=$=

$\frac{\log_qa}{\log_qp}$logq​alogq​p​

Look carefully at this result. It is saying that:

$\log_pa$logp​a

$=$=

$\frac{\log_qa}{\log_qp}$logq​alogq​p​

For example:

$\log_58$log5​8

$=$=

$\frac{\log_{10}8}{\log_{10}5}$log10​8log10​5​

So if we knew the common logs of $8$8 and $5$5, we could determine $\log_58$log5​8. Now, to five decimal places:

$\log_{10}8$log10​8	$=$=	$0.90309$0.90309
$\log_{10}5$log10​5	$=$=	$0.69897$0.69897

Then we have

$\log_58$log5​8	$\approx$≈	$\frac{0.90309}{0.69897}$0.903090.69897​
	$=$=	$1.29203$1.29203 ($5$5 d.p.)

Note that $\log_58$log5​8 could also be expressed as $\frac{\log_2\left(8\right)}{\log_2\left(5\right)}$log2​(8)log2​(5)​ which would give the same answer $1.29203$1.29203. Any base can be used, now we have the above relationship.

 

As another example:

$\log_b10$logb​10	$=$=	$\frac{\log_{10}10}{\log_{10}b}$log10​10log10​b​
$\log_b10$logb​10	$=$=	$\frac{1}{\log_{10}b}$1log10​b​

And:

$\log_b10\times\log_{10}b$logb​10×log10​b

$=$=

$1$1

This is an interesting result. We can generalise this to show that:

$\log_ba\times\log_ab$logb​a×loga​b

$=$=

$1$1

So that $\log_ba$logb​a and $\log_ab$loga​b are multiplicative inverses.

 

 

Practice questions

Question 9

Question 10