# Exponential Function

Function of the form `f(x)=a^x` where `a>0` is called exponential function.

Do not confuse it with power function `f(x)=x^a` in which the variable is the base.

Domain of the exponential function is `(-oo,oo)`, range is `(0,oo)`, provided `a!=1`.

If `a=1` then `f(x)=1^x=1`.

Exponential functions are useful for modeling many natural phenomena, such as population growth (if `a>1`) and radioactive decay (if `0<a<1`).

Let's see what means under `f(x)=a^x`.

If `x=n` where `n` is postive integer then `a^n=\underbrace (a*a*...*a)_{n}`.

If `x=0` then `a^0=1` and if `x=-n` where n is positive integer then `a^(-n)=1/(a^n)`.

If `x` is rational number `x=p/q` where `p` and `q` are integer then `a^x=a^(p/q)=root(q)(a^p)=(root(q)a)^p`.

There are three kinds of exponential functions:

- If `0<a<1` then exponential function decreases.
- If `a=1`, exponential function is constant.
- If `a>1`, it increases (the bigger `a`, the more rapidly it increases).

**Properties of Exponents**

If `a` and `b` are positive numbers, and `x` and `y` are any real numbers, then

- `a^(x+y)=a^xa^y`
- `a^(x-y)=(a^x)/(a^y)`
- `(a^x)^(y)=a^(xy)`

- `(ab)^x=a^xb^x`

**Applications of Exponential Functions**

As was stated above exponential functions are widely used in growth/decay problems.

**Example 1**. Suppose that initial population of some bacteria is 100. It is known that every 1 hours population triples. Find population after 10 hours.

If number of bacteria at time `t` is `p(t)`, where `t` is measured in hours and p(0)=100 is initial population then

`p(1)=3p(0)=300`

`p(2)=3p(1)=3*3*p(0)=3^2p(0)=900`

`p(3)=3p(2)=3*3*3*p(0)=3^3p(0)=2700`

`p(4)=3p(3)=3*3*3*3*p(0)=3^4p(0)=8100`.

Do you see pattern? It seems that `p(t)=100*3^n`.

So, `p(10)=100*3^(10)=5904900`.

**Example 2.** Suppose that half-life of some radioactive element is 3 years (recall that half-life is amount of time needed to disintegrate half of any quantity initially presented). Suppose that initially presented 1000 mg. Find mass after 10 years.

If `m(t)` is mass of element at any time `t` measured in years then

`m(0)=1000`

`m(3)=1/2 m(0)=500`

`m(6)=1/2 m(3)=1/2^2 m(0)=250`.

From this pattern we can conclude that

`m(t)=1/(2^(t/3))*1000=1000*2^(-t/3)`.

So, in 10 years there will be `m(10)=1000*2^(-10/3)~~99.21` mg.