## 2.3 Logarithmic Functions and Models

(This topic is also in Section 2.3 in Applied Calculus and Section 10.3 in Finite Mathematics and Applied Calculus)

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### More on Logarithms

Brush up on your Algebra For all of the tutorials on nonlinear functions and models, you should be familiar with the algebra of exponents and radicals.

Logarithms were invented by John Napier (1550-1617) in the late 16th century as a means of aiding calculation. Although computers and calculators have done away with that use of logarithms, many other uses remain. In particular, the logarithm is used to to model real world phenomena in numerous fields, including physics, finance, and economics.

Base b Logarithm

The base b logarithm of x, logbx, is the power to which we need to raise the positive number b in order to get x.

For instance, the power to which we need to raise 2 in order to get 8 is 3. Therefore,

log28 = 3         because 23 = 8

Filling-In-The-Box-Method of Computing Logarithms
To compute logbx, write down the equation

b = x
and then fill in the box with the correct value. That is the desired logarithm.
Examples
 1.   To compute log39, write 3□ = 9 and fill in the box. 3 2  = 9 Therefore, log39 = 2.
 2.   To compute log2 18 , write 2□ = 18 and fill in the box.
 2-3  = 18
 Therefore, log2 18 = -3.
 3.   To compute log101000, write 10□ = 1000 and fill in the box.
 log101000 =
 4.   More for you to practice with.
 log41 =
 log164 =
 log4 164 =

In Short Writing a logarithmic equation is just a funny way of writing an exponential equation:
 Logarithmic form Exponential form Technology Form logbx = y means by = x log(x,b) = y log525 = 2 52 = 25 log(25,5) = 2 log232 = 5 40.5 = 2 Use technology form

Using Technology

To compute logbx using technology, use the following formulas (see "Change-of-Base Formula" below for an explanation of the first one):

 TI-83: log(x)/log(b) Example: log2(16) is log(16)/log(2) Excel: =LOG(x,b) or =log(x,b) Example: log2(16) is =LOG(16,2)

Common Logarithm, Natural Logarithm

 log10x is usually written as log x Common Logarithm I-83 & Excel Formula: log(x) logex is usually written as ln x Natural Logarithm I-83 & Excel Formula: ln(x)

The next practice question is similar to Example 2 in Section 2.3 of Applied Calculus and Section 10.3 in Finite Mathematics and Applied Calculus.

Let f(x) = log2x. Fill in the missing values in the following table, and then click on the correct graph.

x
 18
 14
 12
1 2 4 16
f(x) = log2x -3

2 4

The (correct) graph above is an example of a logarithmic function. See te textbook for a discussion of general logarithmic functions and logarithmic regression.

One thing that makes logarithms useful for solving equations is their properties:

Logarithmic Identities

The following identities hold for all positive bases a &ne 1 and b &ne 1, all positive numbers x and y, and every real number r. How they are derived is discussed in the on-line discussion "Using and Deriving Algebraic Properties of Logarithms."

Identity Examples
1. logb(xy) = logbx + logby log216 = log28 + log22
2. logb(x/y) = logbx - logby ln(5/3) = ln 5 - ln 3
3. logb(xr) = r logbx log2(65) = 5 log26
4. logbb = 1;     logb1 = 0 log33 = 1;     log31 = 0
 5 logbx = log x log b = ln x ln b
 log26 = log 6 log 2 = ln 6 ln 2
"Change of Base Formula"

Q Multiple Choice: ln(1/2) =

 ln(2) - ln(1) ln(1)/ln(2) 0 undefined e0.5 -ln(2)

Q Multiple Choice: log(18) =

 log(10) + log(8) log(9)log(2) log(10)log(8) 2log(3) + log(2) log(3)2log(2) log(20) - log(2)

Q Multiple Choice: log(0.125) =

 log(10) + log(8) -3log(5) -3log(2) -2log(4) log(125)/3 -5log(3)

Q Multiple Choice: ln[3(1.01-2t)] =

 ln(3) - 2t ln(1.01) -2t(ln(3) + ln(1.01)) 2t(ln(3) - ln(1.01)) ln(3) + ln(1.01) - ln(2) - ln(t) ln(3) - ln(2t) ln(1.01) ln(3) - [ln(2) - ln(t) ln(1.01)]

### Solving Equations with Unknowns in the Exponent

Logarithms are very useful in solving equatinos where the unknown is in the exponent. First go through the sample question and both methods of solution, and then try the others on your own:

Q Sample Question: Solve the equation 52x = 1/125

Solution Method 1 (Rewriting in Log Form):
 Write the equation: 52x = 1/125 Make sure it has the form A = bc : It does. Rewrite in logarithmic form: 2x = log5(1/125) Evaluate: 2x = -3 Solve for x: x = -1.5

Solution Method 2 (Taking the Log of Both Sides):

 Write the equation: 52x = 1/125 Take the log base 5 of both sides: log5(52x) = log5(1/125) = -3 Use rule(3): 2xlog5(5) = -3 Use rule(4): 2x = -3 Solve for x: x = -1.5

Some for you to do:

 Q Solve 4-3x = 64. Answer: x = Q Solve 3(24x) = 192. Answer: x = Q Solve 1000(1.10)4x = 2000 Answer: x = Accurate to 3 decimal places

Relationship with Exponential Functions

The following identities show that the operations of taking the base b logarithm and raising b to a power are "inverse" to each other:

 Identity Examples 1. logb(bx) = x log2(27) = 7 All this says is that The power to which you raise b in order to get bx is x (!) 2. blogbx = x 5log58 = 8 All this says is that raising b to the power to which it must be raised to get x, yields x (!)

### Exponential Growth & Decay and Half-life

At the end of the previous tutorial we saw that exponential growth can be represented by the equation A = Pert, where A, P, and r are constants. Here we generalize this idea (and also change the letters used to represent the constants):

Exponential Growth & Decay

An exponential growth function has the form

 Q(t) = Q0ekt
Q0 represents the value of Q at time t = 0, and k is the growth constant.

Similarly, an exponential decay function has the form

 Q(t) = Q0e-kt
Q0 represents the value of Q at time t = 0, and k is the decay constant.

Of interest here are the following questions:

 If a quantity is undergoing exponential growth, how long does it take for the quantity to double its original size? This time is called the doubling time. If a quantity is undergoing exponential decay, how long does it take for the quantity to decay to half its original size? This time is called the half-life

It turns out that the answers to these questions are independent of the size of the sample: small samples undergoing exponential decay have the same half-life as large samples of the same material, and the same is true for doubling time in the case of exponential growth.

Doubling Time and Half-Life

For a quantity undergoing exponential growth Q(t) = Q0ekt,the growth constant k and doubling time td for Q are related by

 td k = ln 2.

For a quantity undergoing exponential decay Q(t) = Q0e-kt,the decay constant k and half-life th for Q are related by

 th k = ln 2.
Notice that the formulas are exactly the same for both growth and decay. Thus, in both exponential growth and decay,
 td = ln 2k ,      and k = ln 2td
Sample Questions:

Q Carbon-14 has the following decay function: Q(t) = Q0e-0.000 120 968t. Its half-life is therefore

 th = years.

Q If the quantity of a substance doubles every 10 years, then its growth constant is given by
 k = Accurate to 4 decimal places

You now have several options

• Try some of the questions in the true/false quiz (warning: it covers the whole Chapter) by pressing the button on the sidebar.
• Try some of the on-line review exercises (press the "Review Exercises" button on the side.)
• Try some of the exercises from Section 2.3 in Applied Calculus or Section or Section 9.3 of Finite Mathematics and Applied Calculus.

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Last Updated: April, 2007