Exponential Function

growth & decay - Wachstum & Zerfall

We have already come across linear functions and quadratic functions. Some phenomena in nature can not be modelled by these.

example - opening problem

index laws - die Exponentengesetze

Review the index laws (exponents, powers, Hochzahlen, Exponenten)

exponential function - die exponentielle Funktion

example

A bacteria multiplies by binary fission, say once every minute (x).

$$f(x) = 2^x$$

What is the value at x=0? After 6 minutes?

table of values

x 0 1 2 3 4 5 6
${f(x)=2^x}$

growth factor - der Wachstumsfaktor

The base 2 in the example above is called the growth factor $a$. It is the number the amount is multiplied by in a fixed interval (x). The interval is often time.

$$f(x) = a^x$$

growth rate - die Wachstumsrate

The growth rate is a percentage increase or decrease of 100%.

example

Say, a population increases by 3% every year. Then the growth rate is then 3%. To find the growth factor $a$:

$a = 100\% + 3\% = 103\% = 1.03$ as a decimal.

When the population decreases, say by 6%, the growth factor $a$ is calculated as follows.

$a = 100\% - 6\% = 94\% = 0.94$

Note

initial value - der Grundwert

The initial amount is the amount at x = 0. Here, it is denoted by c.

$$f(x) = c \cdot a^x$$

examples

Given $f(x) = 2^x.$

a) find f(0), f(1) and f(4)

Given $f(x) = a^x$. Find the growth factor a if

a) f(1) = 2

b) f(2) = 16

c) f(5) = 243

d) f(10) = 0.45

Given $f(x) = c \cdot 1.3^x$ find c if f(4) = 285.61 .

finding the growth factor

f(n) and f(n+k) given

numerical example

Two points given f(1) = 18 and f(4) = 31.104

$$f(\color{orange}{1}) = 18 = c \cdot a^\color{orange}{1}$$

$$f(\color{lightgreen}{4}) = 31.104 = c \cdot a^{\color{lightgreen}{4}}$$

From the first it follows that

$$c = \frac{18}{a}$$

Substitute $c = \frac{18}{a}$ into the second equation:

$$ 31.104 = \frac{18 \cdot a^{4}}{a} $$

$$ \frac{31.104}{18} = a^{3} $$

Take the 3rd root:

$$ \left(\frac{31.104}{18}\right)^{\frac{1}{3}} = a $$

$$a = \color{purple}{1.2} $$

Substitute a into one of the two original equations to calculate c.

$$ 18 = c \cdot \color{purple}{1.2}^1$$

$$c = \frac{18}{1.2} = \color{orange}{15}$$

So, the equation is

$$f(x) = \color{orange}{15} \cdot \color{purple}{1.2}^x$$

compound interest - der Zinseszins

Let the interest rate (der Zinssatz) be 10% on, say an initial amount is 500€, every year.

Using a formula you may find in books,

$$K = I \cdot (1 + r)^t$$

calculate the capital (das Kapital) after 5 years.

$${\begin{align}K & = 500€ \cdot (1 + 0.1)^t \newline & = 500€ \cdot 1.1^5 \newline & = 805.26€\end{align}}$$

half-life - die Halbwertszeit

Radioactivity is when instable atoms change into a different element by emmitting radiation (alpha-, beta- and gamma-decay). This means the number of atoms gradually decreases over time. When talking about radiactive isotopes, the time it takes for statistically half of the atoms to decay is called the half-life.

$${N_{HL} = N_0 \cdot \frac{1}{2}}$$

To calculate any number of atoms (or proportional mass), following function is used.

$${N_t = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{HL}}}$$

history - reference to physics and the discovery of radioactivity

examples

exponential equations - die Exponentialgleichung

when you have the same base

if you cannot find the same base - the inverse problem

trial and error - guessing the time

logarithms - der Logarithmus

To solve exponential equations for the exponent, you need the inverse function: the logarithm

Using the logarithm to solve an exponential equation for t

$${\begin{align} 145 & = 2.3^t && \text{take log on both sides} \newline \log{145} & = \log{2.3^t} && \text{use log property of powers within the log}\newline \log{145} & = t \cdot \log{2.3} && \text{divide by log 2.3} \newline \frac{\log{145}}{\log{2.3}} & = t && \text{evaluate - swap sides} \newline t & \approx 5.9751 && \end{align}}$$

example with C-14 method

There is 24% of C-14 left in an old animal bone. The half-life of C-14 is 5730a (annum - years). Find the approximate age of the bone.

$${\begin{align} 24 \% & = 100 \% \cdot \left(\frac{1}{2}\right)^\frac{t}{5730} && \text{divide by 100} \newline \frac{24}{100} & = \left(\frac{1}{2}\right)^\frac{t}{5730} && \text{take log on both sides} \newline\log{0.24} & = \log{0.5^\frac{t}{5730}} && \text{use log property of powers within the log}\newline \log{0.24} & = \frac{t}{5730} \cdot \log{0.5} && \text{divide by log 0.5, multiply by 5730} \newline 5730 \cdot \frac{\log{0.24}}{\log{0.5}} & = t && \text{evaluate - swap sides} \newline t & \approx 11797 && \end{align}}$$

The bone is approximately 11800 years old.

keywords


(c) 2019 sebastian.williams[at]sebinberlin.de - impressum und datenschutz - Powered by MathJax & XMin & HUGO & jsxgraph & mypaint