Exponents and logarithms, multiple pages

This is part of our refreshers since these functions are widely used everywhere.

We will just stick to the properties and teach the students how to solve this type of problems.

Exponential functions:
These are functions where $x$ the input variable is an $exponent$ and the base is simply a $constant$ .
For a $base=3$ , if we write: $f(x)=3^{x}$ , we have an exponential function.

Please note that the base $a$ in the following function:
$f(x)=a^{x}$

While $x$ can take the value of any real number, $a$ has restrictions in order to get a valid function:
$a>0$ and $a\neq 1$ .
The graph shows that the function is defined for any value of $x$ and $f(x)$ has a range containing all positive numbers.

Compound interests:
If we have the following information,
$A_0$ =Invested amount (Starting value)
$r$ =annual percentage rate
$n$ = number of times the interest is compounded per year
$t$ = the number of years the money is invested

If $A(t)$ is the amount after t years, we can see:

First year:
Starting $A_0$
For each period the interest rate is the yearly rate over the number of periods.
Amount after 1st period: $A_0+A_0\cdot \frac{r}{n}=A_0(1+\frac{r}{n})$
Amount after 2nd period:
$A_0(1+\frac{r}{n})+(A_0(1+\frac{r}{n}))\cdot (1+\frac{r}{n})=A_0(1+\frac{r}{n})^{2}$

After the $n$ periods: 1 year

$A(1)=A_0(1+\frac{r}{n})^{n\cdot 1}$

For the t years:
$A(t)=A_0(1+\frac{r}{n})^{n\cdot t}$

Example:

Mouctar is investing 1500 dollars in an account that earns 5% interest per year.

What is the amount $A(6)$ , after 6 years if the interest is compounded quarterly.

Solution:

$A(0)=1500$

$r=5\%=0.05$

Per year, $n=4$ , quarterly compounding

Number of years: $t=6$

$A(6)=1500(1+\frac{0.05}{4})^{4\times 6}$

Finally, after 6 years:

$A(6)=2021$ dollars.

Exponential decay functions

When the base of the exponential function is $0<a<1$ , we have a decay.

The models are of the form:

$A(t)=A_0(a)^t$

Example:

A car bought at a price of 12000 dollars is estimated to depreciate such each year the value is only 85% of the previous year.

What will be the value of the car after 5 years?

Solution

We have: $A_0=12000$

$a=0.85$

$t=5$

$A(5)=12000(0.85)^5$

$A(5)=5324.46 \; dollars$

Natural Exponential function

The number $e=2.71828....$ is natural in the following sense.

In our compound interests formula, if we invest $1$ dollar at a 100% interest for one year and we compound continuously, making $n$ very large, we tend to end up with the following amount: