Exponentials [1.5]

Ahhh... remember!

Multiplication of exponentials: $$10^2*10^3=100*1,000=100,000 = 10^{(2+3)}=10^5.$$

Meaning of negative exponentials: $$2^{-3}=\frac{1}{2^3}=\frac18=0.125 .$$

Division of exponentials: (as long as they have the same base!) $$\frac{10^2}{ 10^3}= 100/1000=1/10=0.1=10^{2-3}=10^{-1}=\frac{1}{10}.$$

Exponential raised to a power: $$(3^2)^3=9^3=9\times9\times9=729 = 3^{2*3}=3^6.$$ and in particular... $$e^{kx}= (e^k)^x = (e^x)^k.$$

Useful in problem solving... The second raising "reverses" the effect of the first raising: $$(17.8^3)^{1/3}=17.8^{(3*1/3)}=17.8^1= 17.8.$$ so to solve this for $x$... $$\begineq x^4 &=13\\ (x^4)^{1/4}&=13^{1/4}\\ x &= 13^{0.25}=1.8988 \text{ using a calculator} \endeq$$

[any number]${}^{\color{red}0} =1; \ \ e^0=1; \ \ 17.8^0=1$.

But first...HW [1.1]

The most common issue was on Problem 5

The population of Washington DC grew from 1900 to 1950, stayed approx constant during the 1950's and decreased from about 1960 to 2005. Graph the population as a function of years since 1900.

So, the $x$-axis should run from 0 (=1900) to 105 (=2005)

Exponential functions [1.5]

Exponential compared to linear function

$$\nonumber \color{red}{f(x)=mx+b;}\ \ \ \color{blue}{g(x)=b\cdot m^x}$$

Linear function: $\color{red}{f(x)=mx+b}$

  • Increase $x$ by one,
    and $y$ increases by $\color{red}{m}$.
  • Units of $\color{red}{m}$ are $y$-units/$x$-units.
  • $b$ is the $y$-intercept, and has the same units as $y$.
  • $\color{red}{m}$ can be positive / negative / 0
  • $\color{red}{m}\gt 0$ means the function is increasing.
  • $\color{red}{m}\lt 0$ means the function is decreasing.

Exponential function: $\color{blue}{g(x)=b\cdot m^x}$

  • Increase $x$ by one,
    and $y$ is multiplied by $\color{blue}{m}$.
  • $m$ has no units. Think of it as the multiplier per $x$-unit.
  • $b$ is the $y$-intercept, or "initial value", and has the same units as $y$.
  • $\color{blue}{m}$ can only be positive
  • $\color{blue}{m}\gt 1$ means the function is increasing.
  • $\color{blue}{m}\lt 1$ means the function is decreasing.
  • $m$ is also called the exponential "base".
  • To grow [or shrink] by a decimal percentage $r$: $m=(1+r)$ [or $m=(1-r)$].

    For example, A \$1,000 bank deposit which is growing with 5% annual interest has a balance after $t$ years of $$M(t)=1,000\cdot(1.05)^t.$$

Problem 6

A product costs \$80 today. How much will the product cost in $t$ days if the price is decreased by

  1. \$4 a day
  2. 5% per day

a) $c=80-4t$
b) $c=80(1-0.05)^t=80(0.95)^t$

Problem 20

The number of passengers using a railway fell from 190,205 to 174,989 during a 5-year period. Find the annual percentage decrease over this period.

$N(5)=N(0)m^5$, so
$$\begineq \frac{N(5)}{N(0)}=\frac{174989}{190205} =0.92&= m^5\\ 0.92^{0.2}&=(m^5)^{0.2}\\ 0.983461&=m\\ 1-0.01654&=1+r \endeq $$ So, $r=-0.01654\approx-0.017\Rightarrow$ an average decrease of 1.7% per year.

ConcepTest problem

Sales at a company are changing according to the formula $$S(t) = 1000(0.82)^t,$$ where $S$ is sales in thousands of dollars and $t$ is measured in years. Sales at this company are:
  1. Increasing by 82% per year
  2. Increasing by 82 thousand dollars per year
  3. Decreasing by 82% per year
  4. Decreasing by 82 thousand dollars per year
  5. Increasing by 18% per year
  6. Increasing by 18 thousand dollars per year
  7. Decreasing by 18% per year
  8. Decreasing by 18 thousand dollars per year

$$S(t)=1000(0.82)^t=1000(1-0.18)^t.$$ So, the multiplier, 0.82, is equal to $1-r$, with $r=-0.18$. This means that sales are g. decreasing by 18% per year.

$S(0)=1000$ means that in the starting year ($t=0$) sales were \$1000.

Problem 25

...A problem like this has been on previous exams.

Determine whether each of the following tables of values could correspond to a linear function, an exponential function, or neither. Find formulas for the linear and exponential functions.


For an exponential function, each time you increase the input by one unit, you increase $y$ by the multiplier. So $\frac{y(1)}{y(0)} = m$, $\frac{y(2)}{y(1)}=m$, etc...

(a) Let's see... $\frac{y(1)}{y(0)} = \frac{12.7}{10.5} = 1.21 $, $\frac{y(2)}{y(1)}=\frac{18.9}{12.7}=1.49$. Oh, already the first two ratios are not the same. This is not an exponential function.

(b) Let's see... $\frac{s(0)}{s(-1)} = \frac{50.2}{30.12} = 0.60$, $\frac{s(1)}{s(0)}=\frac{18.072}{30.12}=0.60$ . Keep checking the remaining ratio... $\frac{s(2))}{s(1)}=\frac{10.8432}{18.072}=0.60$. They are all the same, so (b) is an exponential function and the mulitiplier is $m=0.6$. An equation is: $$s(t)=30.12\cdot m^t$$

(c) Each time $u$ increases by 2, $g(u)$ changes by -3, therefore (c) is a linear function, a line with a slope, $m=\frac{\Delta y}{\Delta x}=\frac{-3}{2}=-1.5.$ An equation is: $$g(u)=27 -1.5u$$

Compounding

1000 dollars, compounded annually (once a year) at 12% interest $$A(t)=1000(1+0.12)^t=1000*1.12^t$$

1000 dollars, compounded quarterly (4 times a year) at 12% interest $$A(t)=1000(1+0.12/4)^{4t}=1000[(1.03)^{4}]^t=1000[1.1255]^t$$

12.55% is the Annual Percentage Rate (APR) of this scheme

In general, compounding $n$ times a year with an interest rate of $r$ $$A(t)=A_0(1+r/n)^{nt}.$$

Continuous compounding: $$A(t)=\lim_{n\to \infty} A_0(1+r/n)^{nt}=A_0e^{rt}$$ Where $e$ is Euler's number, an irrational number, approximately equal to 2.71828.