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 $x.$

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$b$:$f(x) \approx \Bigl(\frac{N}{1+A}\Bigr)b^x.$.$b \gt 1b \lt 1$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: