The logarithmic function provides an alternate way to write exponential expressions and is in fact the inverse function of the exponential function.
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.