Natural logarithms are logarithms with the natural base $e$e, where $e\approx2.718281828459$e≈2.718281828459. We previously explored exponential functions with this base and their importance in calculus. Using logarithms with this base will allow us to manipulate and solve applications involving exponentials, further applications in logarithms, as well as apply calculus to general exponential functions.
Natural logarithms are frequently used due to their wide-ranging applications. For this reason, many modern textbooks and calculators abbreviate the notation $\log_ex$logex to $\ln x$lnx.
Applying the definition
Recall from the definition of logarithms that if $y=\log_ex=\ln x$y=logex=lnx, then $x=e^y$x=ey. We can state this in words as '$\ln x$lnx is the index that $e$e must be raised to in order to obtain $x$x'.
Just as with a general base, the functions $f\left(x\right)=e^x$f(x)=ex and $g\left(x\right)=\ln x$g(x)=lnx are inverse functions. 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)) | $=$= | $e^{\ln x}$elnx |
|
| $=$= | $x$x |
Since $\ln x$lnx is the index $e$e 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)) | $=$= | $\ln\left(e^x\right)$ln(ex) |
|
| $=$= | $x$x |
Since we are asking what index $e$e must be raised to obtain $e^x$ex. |
Thus, the functions are inverse operations - they 'undo' each other. This property is particularly useful in solving equations.
$e^{\ln a}=a$elna=a, for $a>0$a>0
$\ln e^a=a$lnea=a
The natural logarithms follow the other logarithm properties, which can be useful for manipulating expressions.
For $x$x and $y>0$y>0, we have:
$\ln x+\ln y=\ln\left(xy\right)$lnx+lny=ln(xy)
$\ln x-\ln y=\ln\left(\frac{x}{y}\right)$lnx−lny=ln(xy)
$n\ln x=\ln\left(x^n\right)$nlnx=ln(xn)
$\ln1=0$ln1=0 and $\ln e=1$lne=1
Worked examples
example 1
Simplify $\frac{\ln\left(x^2\right)}{\ln\left(\sqrt{x}\right)}$ln(x2)ln(√x).
Think: Using the power rule, $n\ln x=\ln\left(x^n\right)$nlnx=ln(xn), we can rewrite the numerator and denominator in terms of $\ln x$lnx and then divide by a common factor.
Do:
| $\frac{\ln\left(x^2\right)}{\ln\left(\sqrt{x}\right)}$ln(x2)ln(√x) | $=$= | $\frac{\ln\left(x^2\right)}{\ln\left(x^{\frac{1}{2}}\right)}$ln(x2)ln(x12) |
Rewrite the square root as a power. |
| $=$= | $\frac{2\ln x}{\frac{1}{2}\ln x}$2lnx12lnx |
Using the power rule to rewrite the terms. |
|
| $=$= | $\frac{2}{\frac{1}{2}}$212 |
Divide the numerator and denominator by $\ln x$lnx. |
|
| $=$= | $4$4 |
Simplify. |
EXAMPLE 2
Solve $e^{2x+5}=3$e2x+5=3.
Think: The unknown $x$x is in the exponent. Taking the logarithm base $e$e of both sides will allow us to bring the expression with the unknown out of the exponent and solve for $x$x.
Do:
| $e^{2x+5}$e2x+5 | $=$= | $3$3 |
|
| $\ln\left(e^{2x+5}\right)$ln(e2x+5) | $=$= | $\ln3$ln3 |
Take the log base $e$e of both sides. |
| $\therefore\ 2x+5$∴ 2x+5 | $=$= | $\ln3$ln3 |
Use the identity $\ln e^a=a$lnea=a. |
| $2x$2x | $=$= | $\ln3-5$ln3−5 |
Rearrange for $x$x. |
| $x$x | $=$= | $\frac{\ln3-5}{2}$ln3−52 |
|
Example 3
Solve $\ln\left(x-2\right)+\ln\left(x+2\right)=\ln5$ln(x−2)+ln(x+2)=ln5.
Think: Combine the two logarithmic terms on the left-hand side to form a single logarithm. Then equate the arguments on both sides to solve for $x$x.
Do:
| $\ln\left(x-2\right)+\ln\left(x+2\right)$ln(x−2)+ln(x+2) | $=$= | $\ln5$ln5 |
|
| $\ln\left(\left(x-2\right)\left(x+2\right)\right)$ln((x−2)(x+2)) | $=$= | $\ln5$ln5 |
Combine the log expressions using $\ln A+\ln B=\ln\left(AB\right)$lnA+lnB=ln(AB). |
| $\ln\left(x^2-4\right)$ln(x2−4) | $=$= | $\ln5$ln5 |
Expand the brackets. |
| $x^2-4$x2−4 | $=$= | $5$5 |
Equate the arguments, that is if $\ln(A)=\ln(B)$ln(A)=ln(B), then $A=B$A=B. |
| $x^2$x2 | $=$= | $9$9 |
Rearrange for $x$x. |
| $x$x | $=$= | $\pm3$±3 |
|
However, the logarithmic terms in the original equation imply that $x>2$x>2, since the argument of a logarithmic expression must be positive. Hence, only the positive case is a viable solution. That is, $x=3$x=3
Example 4
(a) Write $e^{x+\ln9}=2e^{2x}+4$ex+ln9=2e2x+4 in the form $ay^2+by+c=0$ay2+by+c=0, where $y=e^x$y=ex.
Think: First use the index law $A^m×A^n=A^{m+n}$Am×An=Am+n to rewrite $e^{x+\ln9}$ex+ln9. Then rearrange to the given form by using the fact that $e^{2x}=\left(e^x\right)^2$e2x=(ex)2.
| $e^{x+\ln9}$ex+ln9 | $=$= | $2e^{2x}+4$2e2x+4 |
|
| $e^x\times e^{\ln9}$ex×eln9 | $=$= | $2e^{2x}+4$2e2x+4 |
Use the index law $A^m×A^n=A^{m+n}$Am×An=Am+n to rewrite $e^{x+\ln9}$ex+ln9. |
| $9e^x$9ex | $=$= | $2\left(e^x\right)^2+4$2(ex)2+4 |
Use the identity $e^{\ln a}=a$elna=a and $e^{2x}=\left(e^x\right)^2$e2x=(ex)2 to simplify. |
| $2\left(e^x\right)^2-9\left(e^x\right)+4$2(ex)2−9(ex)+4 | $=$= | $0$0 |
Rearrange to the required form by bringing the terms to one side. |
(b) Hence, solve $e^{x+\ln9}=2e^{2x}+4$ex+ln9=2e2x+4.
Think: From part (a) we have $2y^2-9y+4=0$2y2−9y+4=0, where $y=e^x$y=ex. We will first solve the quadratic equation in $y$y, then solve for $x$x.
Do:
| $2y^2-9y+4$2y2−9y+4 | $=$= | $0$0 |
Write out the equation from part (a) in terms of $y$y |
| $\left(2y-1\right)\left(y-4\right)$(2y−1)(y−4) | $=$= | $0$0 |
Factorise. |
| $y$y | $=$= | $\frac{1}{2},4$12,4 |
Solve using the null-factor law. |
| $\therefore\ e^x$∴ ex | $=$= | $\frac{1}{2},4$12,4 |
Replace $y$y with $e^x$ex. |
| $\therefore\ x$∴ x | $=$= | $\ln\left(\frac{1}{2}\right),\ln4$ln(12),ln4 |
Take the log base $e$e of both sides. |
| $=$= | $-\ln2,\ln4$−ln2,ln4 |
|
Both solutions here give positive arguments and are valid.
When solving equations involving logarithms ensure your answers make sense in context and give positive arguments (that is $a>0$a>0 for $\ln a$lna) for any log expressions in the equation or solution.
Practice questions
Example 1
Evaluate $\ln\left(\frac{1}{e^2}\right)$ln(1e2).
Example 2
Solve $\ln\left(\ln e^{-y}\right)=\ln3$ln(lne−y)=ln3.
Example 3
Solve $e^{x+\ln8}=5e^x+3$ex+ln8=5ex+3.
Graphs of natural logarithmic functions
We have seen that $y=e^x$y=ex and $y=\log_ex$y=logex are inverse functions, and as such will be a reflection of one another across the line $y=x$y=x:
Graphs of $y=e^x$y=ex and $y=\log_ex$y=logex
As with the graphs of general logarithmic functions of the form $\log_bx$logbx, we can observe the following properties of the graph of the natural logarithm function $y=\log_ex$y=logex:
- Domain: the argument, $x$x, is restricted to only positive values. That is, $x>0$x>0.
- Range: all real $y$y values can be obtained.
- Asymptotes: there is a vertical asymptote at $x=0$x=0 (on the $y$y-axis). As a result, there is no $y$y-intercept.
- $x$x-intercept: The logarithm of $1$1 is $0$0, irrespective of the base used. As a result, the graph of $y=\log_ex$y=logex intersects the $x$x-axis at $\left(1,0\right)$(1,0).
We can also transform the natural logarithmic function as we transformed the logarithmic function with a general base.
To obtain the graph of $y=a\log_e\left(x-h\right)+k$y=aloge(x−h)+k from the graph of $y=\log_ex$y=logex:
- $a$a dilates (stretches) the graph by a factor of $a$a from the $x$x-axis, parallel to the $y$y-axis
- When $a<0$a<0 the graph is reflected about the $x$x-axis
- $h$h translates the graph $h$h units horizontally, the graph shifts $h$h units to the right when $h>0$h>0 and $\left|h\right|$|h| units to the left when $h<0$h<0
- $k$k translates the graph $k$k units vertically, the graph shifts $k$k units upwards when $k>0$k>0 and $\left|k\right|$|k| units downwards when $k<0$k<0.
Practice questions
Question 4
The functions $y=\log_2x$y=log2x and $y=\log_3x$y=log3x have been graphed on the same coordinate axes.
Using $e=2.718$e=2.718 and by considering the graph of $y=\log_ex$y=logex, complete the statement below:
For $x>\editable{}$x>, the graph of $y=\log_ex$y=logex will lie above the graph of $y=\log_{\editable{}}x$y=logx and below the graph of $y=\log_{\editable{}}x$y=logx.
For $x<\editable{}$x<, the graph of $y=\log_ex$y=logex will lie above the graph of $y=\log_{\editable{}}x$y=logx and below the graph of $y=\log_{\editable{}}x$y=logx.
Question 5
The graphs of $f\left(x\right)=\ln x$f(x)=lnx and $g\left(x\right)$g(x) are shown below.
Which type of transformation is required to get from $f\left(x\right)$f(x) to $g\left(x\right)$g(x)?
Horizontal translation
AReflection
BHorizontal dilation
CVertical translation
DHence state the equation of $g\left(x\right)$g(x).
Question 6
Consider the function $f\left(x\right)=\ln\left(e^2x\right)$f(x)=ln(e2x).
By using logarithmic properties, rewrite $f\left(x\right)$f(x) as a sum in simplified form.
How can the graph of $f\left(x\right)$f(x) be obtained from the graph of $g\left(x\right)=\ln x$g(x)=lnx?
The graph of $f\left(x\right)$f(x) is obtained by stretching the graph of $g\left(x\right)=\ln x$g(x)=lnx vertically by a factor $e^2$e2.
AThe graph of $f\left(x\right)$f(x) is obtained by translating the graph of $g\left(x\right)=\ln x$g(x)=lnx $2$2 units up.
BThe graph of $f\left(x\right)$f(x) is obtained by compressing the graph of $g\left(x\right)=\ln x$g(x)=lnx vertically by a factor $2$2.
CThe graph of $f\left(x\right)$f(x) is obtained by translating the graph of $g\left(x\right)=\ln x$g(x)=lnx $2$2 units to the right.
D