Natural Logarithmic Function
The natural logarithm function \( y = ln(x) \) is the logarithm with the base equal to the Euler Constant e and is therefore the inverse function of the natural exponential function \( y = e^x \). Hence
\[ y = ln(x) \;\;\; \text{if and only if} \;\;\; x = e^y \]
The graphs of \( y = ln(x) \) and its inverse \( y = e^x \) are shown below. Each of the two graphs is a reflection of the second on the line y = x because they are inverse of each other.
In general the composition of a function \( f \) and its inverse \( f^{1} \) are related by:
1) \( f(f^{1}(x)) = x \), x in the domain of \( f^{1} \)
and
2)\( f^{1}(f(x)) = x \), x in the domain of f.
Hence
Since \( ln(x) \) and \( e^x \) are inverse of each other, we have
\[ ln(e^x) = x \]
and
\[ e^{ln(x)} = x , x \gt 0\] .
Numerical examples
\( ln(e^3) = 3\)
\( ln(1) = ln(e^0) = 0 \)
\( ln(e) = ln(e^1) = 1\)
\( ln(1/e) = ln(e^{1}) =  1\)
\( ln(\sqrt e) = ln(e^{1/2}) = 1/2\)
Properties and Rules of the Natural Logarithmic Functions
Domain: \( (0 , +\infty) \)
Range: \( (\infty , +\infty) \)
x inttercept \( (1,0) \)
Vertical Asymptote: \( x = 0\) because \( \lim_{x \to 0^{+}} \ln(x) =  \infty \)
Monotonicity: increasing on the interval \( (0 , +\infty) \)
Continuity: continuous on the interval \( (0 , +\infty) \)
Differentiability: differentiable on the interval \( (0 , +\infty) \)
One to One Function: It is a one to one function
Inverse Function: The inverse of the natural logarithm function \( f(x) = ln(x) \) is \( g(x) = e^x) \)
Composition with Inverse: \( \ln(e^x) = x \) and \( e^{\ln(x)} = x \) for \( x \gt 0 \)
Derivative: \( \dfrac{d \; ln(x)}{dx} = \dfrac{1}{x} \)
Rules of the natural logarithm ln(x)
\( ln(a \times b) = ln(a) + ln(b) \) , for a anb b positive
\( ln(\dfrac{a}{b}) = ln(a)  ln(b) \) , for a anb b positive
\( n \; ln(x) = ln(x^n) \) , for x positive
A logarithm to any base may be written as a natural logarithm using the change of base formula
\( log_a(x) = \dfrac{ln(x)}{ln(a)} \)
Mathematical Models Using Exponential Function
Natural logarithmic functions are used to solve equations related to mathematical models including natural exponential functions.
Example 1: Population Growhth Modeling
The population P of a small city varies continously according to the formula
\[ P = 10000 e^{0.025 t} \]
where t is the number of years after the year 2019.
When will be the popultaion reach 12000?
Solution to Example 1
We know the population and we want to find t.
Hence the equation
\( 12000 = 10000 e^{0.025 t}\)
Divide both sides of the equation by 10000 and simplify
\( 1.2 = e^{0.025 t}\)
take the natural logarithm of both sides
\( ln(1.2) = ln(e^{0.025 t}) \)
Use the property \( ln(e^x) = x \) to simplify the right side of the equation
\( ln(1.2) = 0.025 t \)
Divide both sides of the equation by 0.025 and evaluate t
\( t = ln(1.2) / 0.025 = 7.29286 years\)
We need to round t = 7.29286 years to the nearest integer, hence
\( 2019 + 7 = 2026\)
The population of the city will reach 12000 in the year 2026.
Example 2: Continuous Compounding Interest
The balance B in a saving account continously compounded is given by
\[ B = P e^{r t}\]
where r is the annual rate of interest and \( P \) is the principal.
An amout of \( \$ 50000 \) is invested at the rate \( r = 5.5\% \) and compounded continuously. How many years, after the initial deposit of \( \$50000\), will it take for the investment to reach \( \$75000 \)?
Solution to Example 2
Let t = 0 when the initial \( \$50000 \) is deposited. Hence the balance B is given by
\[ B = 50000 e^{0.055 t}\]
we need to find t when B is equal to $75000.
\( 75000 = 50000 e^{0.055 t} \)
divide both sides of the equation by 50000 and simplify to obtain
\( 1.5 = e^{0.055 t} \)
take natural logarithm of both sides
\( ln( 1.5 ) = ln (e^{0.055 t}) \)
Use the property \( \ln(e^x) = x \) to simplify right side of the equation and
\( 0.055 t = ln(1.5) \)
\( t = ln(1.5) / 0.055 \approx 7.4 \) years
