Exploring the inverse relationship
The functions $f\left(x\right)=b^x$f(x)=bx and $g\left(x\right)=\log_bx$g(x)=logbx are inverse functions. This means the functions 'reverse' one another, so applying one and then the other in either order will produce no net effect. We have already seen inverse functions before - think about what happens if we multiply a number by $2$2 and then apply the inverse operation of dividing by $2$2.
We can observe this property 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 power $b$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 power $b$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.
$b^{\log_bx}=x$blogbx=x, for $x>0$x>0 and $b>0$b>0, $b\ne1$b≠1
$\log_bb^x=x$logbbx=x, for $b>0$b>0, $b\ne1$b≠1
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.
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:
$\log_bx-\log_by=\log_b\left(\frac{x}{y}\right)$logbx−logby=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 |
| $=$= | $b^{N-M}$bN−M |
Using the index law for dividing exponential terms with the same base |
|
| $\log_b\left(\frac{x}{y}\right)$logb(xy) | $=$= | $N-M$N−M |
Rewriting using logarithms |
| $=$= | $\log_bx-\log_by$logbx−logby |
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 $\log_310-\log_32$log310−log32.
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:
| $\log_310-\log_32$log310−log32 | $=$= | $\log_3\left(\frac{10}{2}\right)$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.
$\log_a\left(x^n\right)=n\log_ax$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. Which can also 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. Which can also be interpreted as "the power of $a$a that obtains $a$a is one".
$\log_a1=0$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(x−5) |
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$5log2p−log2p3+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$5log2p−log2p3+3 | $=$= | $5\log_2p-\log_2p^3+3\log_22$5log2p−log2p3+3log22 |
Using the identity $\log_aa=1$logaa=1, we can introduce a log into our constant value |
| $=$= | $\log_2p^5-\log_2p^3+\log_22^3$log2p5−log2p3+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(3√116).
question 4
Express $7\log x-\log\left(\frac{1}{x}\right)-\log y$7logx−log(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.
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, $a\ne1$a≠1.
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.