A function of the form f(x)=ax, where a>0, is called an exponential function.
Do not confuse it with the power function f(x)=xa in which the variable is the base.
The domain of the exponential function is (−∞,∞), and the range is (0,∞), provided a=1.
If a=1, we have that f(x)=1x=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 is meant under f(x)=ax.
If x=n, where n is a postive integer, we have that an=n(a⋅a⋅…⋅a).
If x=0, it can be stated that a0=1, and if x=−n, where n is a positive integer, we have that a−n=an1.
If x is a rational number, x=qp, where p and q are integers, it can be stated that ax=aqp=qap=(qa)p.
There are three kinds of exponential functions:
If 0<a<1, the exponential function decreases.
If a=1, the 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, it can be stated that:
ax+y=axay
ax−y=ayax
(ax)y=axy
(ab)x=axbx
Applications of exponential functions.
As stated above, exponential functions are widely used in growth/decay problems.
Example 1. Suppose that the initial population of some bacteria is 100. It is known that every 1 hour the population triples. Find the population after 10 hours.
If the number of bacteria at a time t is p(t), where t is measured in hours, and p(0)=100 is the initial population, we have that
p(1)=3p(0)=300
p(2)=3p(1)=3⋅3⋅p(0)=32p(0)=900
p(3)=3p(2)=3⋅3⋅3⋅p(0)=33p(0)=2700
p(4)=3p(3)=3⋅3⋅3⋅3⋅p(0)=34p(0)=8100.
Do you see the pattern? It seems that p(t)=100⋅3n.
So, p(10)=100⋅310=5904900.
Now, let's proceed to another example.
Example 2. Suppose that the half-life of some radioactive element is 3 years (recall that the half-life is the amount of time needed to disintegrate a half of any quantity initially presented). Suppose that the initially presented mass is 1000 mg. Find the mass after 10 years.
If m(t) is the mass of the element at any time t measured in years, we have that
m(0)=1000
m(3)=21m(0)=500
m(6)=21m(3)=221m(0)=250
From this pattern, we can conclude that
m(t)=23t1⋅1000=1000⋅2−3t.
So, in 10 years, there will be m(10)=1000⋅2−310≈99.21 mg.