We are already be familiar with exponential functions. In this chapter, our focus will be on a closely related function called the logarithmic function. The logarithmic function provides an alternate way to write exponential expressions and is in fact the inverse function of the exponential function (we will explore the idea of inverse functions further in our next lesson).
Given an exponential equation of the form:
$b^x=a$bx=a,
This is equivalent to a logarithmic equation of the form:
$\log_ba=x$logba=x.
In this equation, $b$b is called the base, just as it is with an exponential; $a$a is called the argument, and $x$x is the result (the index from the exponential equation).
Note: We read logarithms such as $\log_28$log28 as "the log base $2$2 of $8$8".
A logarithm of the form $\log_ba$logba is equal to the index to which we would raise the base $b$b in order to obtain the argument $a$a.
The table below shows some examples of converting between exponential form and logarithmic form:
| Exponential | Logarithm |
|---|---|
| $2^3=8$23=8 | $\log_28=3$log28=3 |
| $5^{-2}=\frac{1}{25}$5−2=125 | $\log_5\frac{1}{25}=-2$log5125=−2 |
| $10^0=1$100=1 | $\log_{10}1=0$log101=0 |
For a base $b>0$b>0 the result of the exponential expression $b^x$bx for any real $x$x is a positive number. Correspondingly, this means that we can only take the logarithm of a positive argument. We also only talk about logarithms with a positive base (think about why this might be the case). Additionally, since $1^x=1$1x=1 for any $x$x, we cannot take a logarithm with a base of $1$1 (an inverse doesn't make sense here).
In summary:
$\log_ba$logba only makes sense when $a>0$a>0 and $b>0$b>0, $b\ne1$b≠1
For example, we cannot evaluate $\log_2(-4)$log2(−4) nor $\log_{\left(-3\right)}27$log(−3)27.
Worked examples
Example 1
Rewrite $6^2=36$62=36 in logarithmic form.
Think: In this exponential expression we can identify the base as $6$6, the index as $2$2 and the result as $36$36. To transpose it into a logarithmic expression, $\log_ba=x$logba=x, $b$b is the base, $a$a the argument is the result of the exponential equation, and $x$x is the index from the exponential equation.
Do: Thus we have $\log_636=2$log636=2. Which we can read as "the index that $6$6 must be raised to in order to obtain $36$36 is $2$2"
Example 2
Rewrite $\log_{\frac{1}{4}}\frac{1}{64}=3$log14164=3 in exponential form.
Think: From this logarithmic expression we can identify the components of the exponential expression: the base $\frac{1}{4}$14, the index $3$3 and the result $\frac{1}{64}$164.
Do: Thus we have $\left(\frac{1}{4}\right)^3=\frac{1}{64}$(14)3=164
Base $10$10 logarithms, are also known as common logarithms and were originally looked up in a table of logs, prior to being able to evaluate them using calculators. Since they are such a commonly used base, the notation for them is often simplified, with $\log_{10}a$log10a sometimes written simply as $\log a$loga (when the base is unambiguous).
Practice questions
question 1
Rewrite the equation $25^{1.5}=125$251.5=125 in logarithmic form (with the index as the subject of the equation).
question 2
Rewrite $\log_g5=3$logg5=3 in exponential form.
Evaluating logarithmic expressions
Recall that the log expression $\log_ba$logba is the number that the base $b$b must be raised to in order to result in $a$a. This is a key understanding of the expression.
For example, to evaluate $\log_{10}1000$log101000, we ask ourselves "to what index must $10$10 be raised to get a result of $1000$1000?" Since $10$10 raised to the index $3$3 would result $1000$1000, then $3$3 is said to be the base $10$10 logarithm of $1000$1000. That is $\log_{10}1000=3$log101000=3
Consider the square root function, where we can think of the expression $\sqrt{a}$√a both as an operation and a number in its own right. For example $\sqrt{4}$√4 is the positive number which, when multiplied by itself, will obtain $4$4. We can simplify the expression by evaluating the square root, since $\sqrt{4}=2$√4=2. However, some expressions cannot be simplified $\sqrt{5}$√5 is a number when multiplied by itself obtains $5$5 but while we could obtain an approximation using our calculator we cannot express the number exactly without use of the $\sqrt{\ }$√ symbol.
Similarly a logarithmic expression can be considered both an operation or number in its own right. For example $\log_28$log28 is the index $2$2 must be raised to in order to get a result of $8$8. We can simplify the expression by evaluating the logarithm, since $2^3=8$23=8, $\log_28=3$log28=3.
Some expressions cannot be simplified though - for instance, $\log_310$log310 is the index $3$3 must be raised in order to get a result of $10$10. What would that be? Since $3^2=9$32=9 and $3^3=27$33=27 it must be a little larger than $2$2 and we could obtain an approximation using our calculator $\log_310\approx2.0959$log310≈2.0959. If we want to express the number exactly, we would simply leave it as $\log_310$log310.
Worked example
question 3
Evaluate without the use of technology $\log_5125$log5125.
Think: What index must $5$5 be raised to get a result of $125$125? It can be helpful to rewrite $125$125 as a power of $5$5 first, then write down the index.
Do:
| $\log_5125$log5125 | $=$= | $\log_55^3$log553 |
Rewrite $125$125 as a power of $5$5. |
| $=$= | $3$3 |
Simplify by just writing the index. |
Alternatively, we could rewrite the question using a pronumeral as the result, then transpose into an exponential expression and solve for $x$x.
Do: $\log_5125=x$log5125=x
| $5^x$5x | $=$= | $125$125 |
Transposing into an exponential expression. |
| $5^x$5x | $=$= | $5^3$53 |
Making the bases the same on each side. |
| $x$x | $=$= | $3$3 |
Equating the indices. |
Hence $\log_5125=3$log5125=3.
Practice questions
question 3
Evaluate $\log_2\left(\frac{1}{2}\right)$log2(12), without the use of a calculator.
Question 4
Consider the expression $\log8892$log8892.
Within what range is the value of $\log8892$log8892?
between $0$0 and $1$1
Abetween $4$4 and $5$5
Bbetween $2$2 and $3$3
Cbetween $3$3 and $4$4
Dbetween $1$1 and $2$2
EFind the value of $\log8892$log8892 correct to four decimal places.
Question 5
Evaluate $\log_36+2\log_310-5\log_315$log36+2log310−5log315 using technology.
Round your answer to two decimal places.