# 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:

1. If 0<a<1 then exponential function decreases.
2. If a=1, exponential function is constant.
3. 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

1. a^(x+y)=a^xa^y
2. a^(x-y)=(a^x)/(a^y)
3. (a^x)^(y)=a^(xy)
1. (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.