← Part 1
Next topic →
Tutorial
Exercises
Textbooks
    Webmaster
Español
← Part 1Next topic →TutorialExercisesTextbooks     WebmasterEspañol

finite mathematics & applied calculus topic summary:
Nonlinear functions and models: Part 2 of 2
Tools:


Relationship of the functions f(x) = logax and g(x) = ax

If $a$ is any positive number, then the functions $f(x) = \log_ax$ and $g(x) = a^x$ are inverse functions.

This means that
    $a^{\log_ax} = x$         and         $\log_a(a^x) = x$
for every real number $x.$

Want to learn more about inverse functions? Go to the online text: Inverse functions.
Example: Relationship of the functions f(x) = logax and g(x) = ax

$2^{\log_2x} = x$$\log_2(2^x) = x$
$e^{\ln x} = x$$\ln (e^x) = x$

 

Exponential growth

An exponential growth model has the form
    $Q(t) = Q_0e^{kt} \qquad$ (k, Q0 both positive)
$Q_0$ represents the value of $Q$ at time $t = 0$, and $k$ is the growth constant. The doubling time $t_d$ of a substance undergoing exponential growth is the amount of time it takes for the original quantity to double. The doubling time does not depend on the original quantity $Q_0$ of substance.

The growth constant $k$ and doubling time $t_d$ are related by the equation
    $t_dk = \ln 2.$
Examples: Exponential growth

$P(t) = 10\,000e^{0.5t}$ is the value of an account after $t$ years if \$10.000 invested at 5% annually with interest compounded continuously.


If $k = 0.0123$ per year, then $t_d(0.0123) = \ln 2,$ so the doubling time is $t_d = \frac{\ln 2}{k} = \frac{\ln 2}{0.0123} \approx 56.35$ years.


Practice:

 

Exponential decay

An exponential decay model has the form
    $Q(t) = Q_0e^{-kt} \qquad$ (k, Q0 both positive)
$Q_0$ represents the value of $Q$ at time $t = 0$, and $k$ is the decay constant. The half-life $t_d$ of a substance undergoing exponential decay is the amount of time it takes for half the original quantity to decay. As with the doubling time for exponential growth, the half-life does not depend on the original quantity $Q_0$ of substance.

The decay constant $k$ and half-life $t_d$ are related by the equation
    $t_hk = \ln 2.$
Examples: Exponential decay

$Q(t) = Q_0e^{-0,000120968t}$ is the decay function for carbon-14.


If $k = 0.0123$ per year, then $t_h(0.0123) = \ln 2,$ so the half-life is $t_h = \frac{\ln 2}{k} = \frac{\ln 2}{0.0123} \approx 56.35$ years.


Practice:

 

Logistic functions

A logistic function has the form
    $f(x) = \frac{N}{1 + Ab^{-x}}$
for given constants $A$, $N$, and $b$ ($b > 0$ and $b \neq 1$).

Properties of the logistic curve
  • The graph is an S-shaped curve sandwiched between the horizontal lines $y = 0$ and $y = N.$ $N$ is called the limiting value of the logistic curve.
  • If $b \gt 1$, the graph rises; if $b \lt 1,$ the graph falls.
  • The $y$-intercept is $\frac{N}{1 + A}$.
  • Role of $b$: For small values of $x$, the function is approximately exponential with base $b$:
      $f(x) \approx \Bigl(\frac{N}{1+A}\Bigr)b^x.$
    .

$b \gt 1$

$b \lt 1$
Examples: Logistic functions

$N = 50, A = 24, b = 3$ gives
    $f(x) = \frac{N}{1+Ab^{-x}}$ Technology format: 50/(1+24*3^(-x))
The following figure shows the graph of $f$ together with the exponential approximation:

Curva logística: 50/(1+24*3^(-x))
Curva exponencial: 2*3^x


Practice: