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VCE 12 Methods 2023

4.05 Properties of logarithms

Lesson

Exploring the inverse relationship

As mentioned in the previous lesson the functions $f\left(x\right)=b^x$f(x)=bx and $g\left(x\right)=\log_bx$g(x)=logbx are inverse functions. 

We can observe this by substituting one function into the other:

  • Substituting $g\left(x\right)$g(x) into $f\left(x\right)$f(x):
$f\left(g\left(x\right)\right)$f(g(x)) $=$= $b^{\log_bx}$blogbx

 

  $=$= $x$x

Since $\log_bx$logbx is the index $b$b should be raised to in order to obtain $x$x.

 

  • Substituting $f\left(x\right)$f(x) into $g\left(x\right)$g(x):
$g\left(f\left(x\right)\right)$g(f(x)) $=$= $\log_b\left(b^x\right)$logb(bx)

 

  $=$= $x$x

Since we are asking what index $b$b must be raised to obtain $b^x$bx

 

Thus, the functions are inverse operations - they 'undo' each other. This property is particularly useful in solving equations.

Inverse property

$b^{\log_bx}=x$blogbx=x, for $x>0$x>0 and $b>0$b>0, $b\ne1$b1

$\log_bb^x=x$logbbx=x, for $b>0$b>0, $b\ne1$b1

 

Practice question

Question 1

Evaluate $2^{\log_23}$2log23.

Question 2

Given that $a>1$a>1, fully simplify the expression $\log_a\left(\frac{1}{a^2}\right)$loga(1a2).

 

Addition and subtraction property

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 the index law for multiplying exponential terms with the same base

$\log_b\left(xy\right)$logb(xy) $=$= $N+M$N+M

Rewriting using logarithms

  $=$= $\log_bx+\log_by$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 the index law for dividing exponential terms with the same base

$\log_b\left(\frac{x}{y}\right)$logb(xy) $=$= $N-M$NM

Rewriting using logarithms

  $=$= $\log_bx-\log_by$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 $\log_5\left(6x\right)$log5(6x) 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 $\log_5\left(6x\right)$log5(6x) in the form:

$\log_56+\log_5x$log56+log5x

 

Power 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.

 

Notice the power came to the front of the expression. We can extend this to logarithms with other powers using repeated application of the addition property, this property can be summarised as follows for any power $n$n.

Power property of logarithms

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

 

Proof of the power property

Start by letting $\log_bx=m$logbx=m and thus, an equivalent exponential statement is $x=b^m$x=bm. Raising the expression to the power of $n$n gives us the result:

$x^n$xn $=$= $\left(b^m\right)^n$(bm)n

Raise both sides of $x=b^m$x=bm to the power $n$n

$x^n$xn $=$= $b^{mn}$bmn

Use the index law $\left(b^m\right)^n=b^{\left(mn\right)}$(bm)n=b(mn)

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

Express as a logarithm

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

Substitute back for $m=\log_bx$m=logbx

 

Worked example

example 3

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

 

Let's explore two further properties.

From our index laws, we know that $a^0=1$a0=1. By transposing this into a logarithmic expression we can see we have $\log_a1=0$loga1=0. This can be interpreted as "the power of $a$a that obtains $1$1 is zero".

Similarly, we know that $a^1=a$a1=a. By transposing this into a logarithmic expression we can see we have $\log_aa=1$logaa=1. This can be interpreted as "the power of $a$a that obtains $a$a is one".

Two further logarithm identities

$\log_a1=0$loga1=0

$\log_aa=1$logaa=1

 

Worked examples

example 4

Simplify $\frac{\log_3256}{\log_364}$log3256log364

Think: When simplifying a logarithmic expression like this, we first consider whether we can write $256$256 and $64$64 as powers of the base of the logarithm, in this case $3$3. If we can't, instead we can firstly rewrite $256$256 and $64$64 using the same base, and then use the power property to simplify further.

Do:

$\frac{\log_3256}{\log_364}$log3256log364 $=$= $\frac{\log_34^4}{\log_34^3}$log344log343

Rewriting both $256$256 and $64$64 with base $4$4

  $=$= $\frac{4\log_34}{3\log_34}$4log343log34

Using the power property

  $=$= $\frac{4}{3}$43

Simplifying by dividing the numerator and denominator by $\log_34$log34

 

example 5

Evaluate $\log_x(\frac{1}{x^5})$logx(1x5)

Think: With this expression we can see that we can express $\frac{1}{x^5}$1x5 as a power of the base of the logarithm, enabling us to use the power property and one of our two identities.

Do:

$\log_x(\frac{1}{x^5})$logx(1x5) $=$= $\log_x(x^{-5})$logx(x5)

Rewriting $\frac{1}{x^5}$1x5

  $=$= $-5\log_xx$5logxx

Using the power property

  $=$= $-5$5

Using the identity $\log_aa=1$logaa=1

 

example 6

Express $5\log_2p-\log_2p^3+3$5log2plog2p3+3 as a single logarithm.

Think: Using a combination of the rules and identities we have established in this lesson, we can simplify this expression into a single logarithm.

Do:

$5\log_2p-\log_2p^3+3$5log2plog2p3+3 $=$= $5\log_2p-\log_2p^3+3\log_22$5log2plog2p3+3log22

Using the identity $\log_aa=1$logaa=1, we can introduce a log into our constant value

  $=$= $\log_2p^5-\log_2p^3+\log_22^3$log2p5log2p3+log223

Using the power rule to simplify

  $=$= $\log_2(\frac{p^5}{p^3})+\log_28$log2(p5p3)+log28

Using the subtraction property to simplify

  $=$= $\log_2p^2+\log_28$log2p2+log28

Simplifying using index laws

  $=$= $\log_2(8p^2)$log2(8p2)

Using the addition property to simplify

 

Practice questions

Question 3

Evaluate $\log_2\left(\sqrt[3]{\frac{1}{16}}\right)$log2(3116).

question 4

Express $7\log x-\log\left(\frac{1}{x}\right)-\log y$7logxlog(1x)logy as a single $\log$log expression.

question 5

Express $\log\left(\sqrt{\frac{c^8}{d}}\right)$log(c8d) in the form $k\log c+m\log d$klogc+mlogd.

question 6

Using the rounded values $\log_{10}3=0.477$log103=0.477 and $\log_{10}25=1.398$log1025=1.398, calculate $\log_{10}\left(\frac{5}{3}\right)$log10(53) to three decimal places.

 

Change of base

Changing from one base to another has been useful in the past when we had less flexible calculators. Now however, it is most useful to us when we might need to simplify a logarithmic expression in a situation involving calculus.

Change of base

If we have $\log_mn$logmn, this can be rewritten with a change of base as $\frac{\log_an}{\log_am}$loganlogam where $a>0$a>0, $a\ne1$a1.

 

Worked example

example 7

Express $\log_35$log35 in base $4$4.

Think: We can apply the change of base rule, where the new base will be $4$4

Do: $\frac{\log_45}{\log_43}$log45log43

 

Practice questions

Question 7

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

question 8

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

Outcomes

U34.AoS2.11

apply algebraic, logarithmic and circular function properties to the simplification of expressions and the solution of equations

U34.AoS2.7

exponent laws and logarithm laws

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