The natural logarithmic function \[ y = \ln(x) \] is the logarithm whose base is the Euler constant \( e \). It is therefore the inverse function of the natural exponential function \[ y = e^x. \] Hence,
\[ y = \ln(x) \quad \text{if and only if} \quad x = e^y. \]The graphs of \( y = \ln(x) \) and its inverse \( y = e^x \) are shown below. Each graph is the reflection of the other across the line \( y = x \), since they are inverse functions.
In general, the composition of a function \( f \) and its inverse \( f^{-1} \) satisfies:
1) \( f\!\left(f^{-1}(x)\right) = x \), for \( x \) in the domain of \( f^{-1} \)
2) \( f^{-1}(f(x)) = x \), for \( x \) in the domain of \( f \)
Since \( \ln(x) \) and \( e^x \) are inverse functions, we obtain:
\[ \ln\!\left(e^x\right) = x \] and \[ e^{\ln(x)} = x, \quad x > 0. \]Numerical examples
\( \ln(e^3) = 3 \)
\( \ln(1) = \ln(e^0) = 0 \)
\( \ln(e) = \ln(e^1) = 1 \)
\( \ln\!\left(\frac{1}{e}\right) = \ln(e^{-1}) = -1 \)
\( \ln(\sqrt{e}) = \ln(e^{1/2}) = \frac{1}{2} \)
Domain: \( (0, +\infty) \)
Range: \( (-\infty, +\infty) \)
\(x\)-intercept: \( (1, 0) \)
Vertical asymptote: \( x = 0 \), since
\( \displaystyle \lim_{x \to 0^{+}} \ln(x) = -\infty \)
Monotonicity: Increasing on \( (0, +\infty) \)
Continuity: Continuous on \( (0, +\infty) \)
Differentiability: Differentiable on \( (0, +\infty) \)
One-to-one: Yes
Inverse function: If \( f(x) = \ln(x) \), then \( f^{-1}(x) = e^x \)
Composition: \( \ln(e^x) = x \) and \( e^{\ln(x)} = x \) for \( x > 0 \)
Derivative:
\[
\frac{d}{dx}\bigl[\ln(x)\bigr] = \frac{1}{x}
\]
Rules of the natural logarithm
\[ \ln(ab) = \ln(a) + \ln(b), \quad a>0,\; b>0 \] \[ \ln\!\left(\frac{a}{b}\right) = \ln(a) - \ln(b), \quad a>0,\; b>0 \] \[ \ln(x^n) = n\,\ln(x), \quad x>0 \]
Any logarithm can be written in terms of the natural logarithm using the change-of-base formula:
\[ \log_a(x) = \frac{\ln(x)}{\ln(a)}. \]Natural logarithmic functions are commonly used to solve equations involving exponential models.
The population \( P \) of a small city varies continuously according to
\[ P = 10000\,e^{0.025t}, \]where \( t \) is the number of years after 2019. When will the population reach 12 000?
Solution
Set \( P = 12000 \):
\[ 12000 = 10000\,e^{0.025t} \] \[ 1.2 = e^{0.025t} \] \[ \ln(1.2) = 0.025t \] \[ t = \frac{\ln(1.2)}{0.025} \approx 7.29 \text{ years} \]Rounding to the nearest year: \[ 2019 + 7 = 2026. \] The population will reach 12 000 in the year 2026.
For an account with continuous compounding, the balance \( B \) is given by
\[ B = Pe^{rt}, \]where \( P \) is the principal and \( r \) is the annual interest rate. An amount of \( \$50{,}000 \) is invested at \( r = 5.5\% \). How long will it take for the investment to reach \( \$75{,}000 \)?
Solution
\[ 75000 = 50000\,e^{0.055t} \] \[ 1.5 = e^{0.055t} \] \[ \ln(1.5) = 0.055t \] \[ t = \frac{\ln(1.5)}{0.055} \approx 7.4 \text{ years} \]