Section 5-3:
Exponential
Functions

Demo:
Exponential Applet (Kennesaw State University)
*Any function in the
form
f(x) = ab*^{x},
where
a > 0, b > 0 and b not equal to 1 is called an exponential
function with base b. Let's take a look
at
a couple of simple exponential graphs.
*f(x) = 2*^{x}
x |
f(x) |

3 |
8 |

2 |
4 |

1 |
2 |

0 |
1 |

-1 |
1/2 |

-2 |
1/4 |

-3 |
1/8 |

* *
*Notice the domain
is
all real numbers and the range is y > 0. As x gets larger
(right),
y gets very large. As x gets smaller(left), y approaches zero
asymptotically.
Notice also that the graph crosses the y-axis at (0, 1). The
above
is the general shape
of an exponential with b > 1.
This is an example of exponential growth.*

* Now let's
look
at the graph of*
*f(x) = (1/2)*^{x}
x |
f(x) |

3 |
1/8 |

2 |
1/4 |

1 |
1/2 |

0 |
1 |

-1 |
2 |

-2 |
4 |

-3 |
8 |

* *
*Observe that
this
graph is the reflection about the y-axis of the first graph. The
domain is still all real numbers and the range is y > 0. The
y-intercept
is (0, 1). This is the general form of an exponential graph if 0
< b < 1. It is an example of an **exponential
decay.*

* Look at the
following
graphs that illustrate the general properties of exponentials.*
*Do you see the
similarities
of each graph?*
*How about this one?*

Many of the functions
associated
with exponential growth or decay are functions of time. We have
already
had one form:
A(t) = A_{o}(1 +
r)^{t}
A second form looks like:
A(t) = A_{o}b^{t/k}
where k = time needed to
multiply
A_{o} by b

__Rule of 72__
* *
*If a quantity is
growing
at r% per year then the doubling time is approximately 72/r years.*

For example, if a
quantity
grows at 10% per year, then it will take 72/10 or 7.2 years to double
in
value. In other words, it will take you 7.2 years to double your
money if you put it into an account that pays 10% interest. At
the
current bank rate or 2%, it will take you 72/2 or 36 years to double
your
money!! Boy, jump all over that investment!!

__Sample Problems__
* *

*1) Suppose you invest
money so that it grows at A(t) = 1000(2)*^{t/8 }

*a) How much money did you invest?*

*
b) How long will it take to double your money?*

* *

*
Solutions:*

*
a) The original amount in the formula is $1000.*

*
b) This means what time will it take to get $2000.*
*2000 = 1000(2)*^{t/8}
*2 = 2*^{t/8}
*1 = t/8*
*8 = t*
*It will take 8
years
to double your money!!*
* *

*2) Suppose that t
hours
from now the population of a bacteria colony is given by: P(t) =
150(100)*^{t/10}

*
a) What is the initial population?*

*
b) How long does it take for the population to be multiplied by*

*
100?*

*
c) What is the population at t = 20?*

* *

*
Solutions:*

*
a) It is 150 from the original equation.*

*
b) It takes 10 hours. That's the definition of the
exponential*

*
function.*

*
c) P(20) = 150(100)*^{20/10} = 150(100)^{2} =
1,500,000

* *

*3) The half life of
a substance is 5 days. We have 4 kg present now.*

*
a) Write a formula for this decay problem.*

*
b) How much is left after 10 days? 15 days? 20 days?*

* *

*
Solutions:*

*
a) A(t) = 4(1/2)*^{t/5}

*
b) A(10) = 4(1/2)*^{10/5} = 4(1/2)^{2} = 1 kg.

*
A(15) = 4(1/2)*^{15/5} = 4(1/2)^{3} = 1/2 kg.

*
A(20) = 4(1/2)*^{20/5} = 4(1/2)^{4} = 1/4 kg.

* *

*4) The value of a car
is given by the equation V(t) = 6000(.82)*^{t}

*
a) What is the annual rate of depreciation?*

*
b) What is the current value?*

*
c) What will be the value in three years?*

* *

*
Solutions:*

*
a) It is 1 - .82 or .18 = 18%*

*
b) The current value is given in the formula, $6000.*

*
c) V(3) = 6000(.82)*^{3} = 3308.21 Which is $3308.21

*On to the number
e!! *
* *
*Back up, I'm
lost!! *
* *