**Relationship of the functions ***f*(*x*) = log_{a}x and *g*(*x*) = *a*^{x}
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*) = log_{a}x and *g*(*x*) = *a*^{x}
$2^{\log_2x} = x$ | | $\log_2(2^x) = x$ |

$e^{\ln x} = x$ | | $\ln (e^x) = x$ |

**Exponential growth
**
An

**exponential growth model** has the form

$Q(t) = Q_0e^{kt} \qquad$ (*k*, *Q*_{0} 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

**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*, *Q*_{0} 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

**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:**