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.
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 power 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 power $e$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 power $e$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 also 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
question 1
Evaluate $\ln\left(\frac{1}{e^2}\right)$ln(1e2).
question 2
Solve $\ln\left(\ln e^{-y}\right)=\ln3$ln(lne−y)=ln3.
question 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