Using Logarithms

Class Notes Precalculus I, 10/29/98

Using Logarithms

Today's quiz problem was to make a partial table of y = 5 ^ x, then use this table to construct a partial table for y = log{base 5} (x).

The table for y = 5 ^ x is constructed below. The table for y = log{base 5}(x) is obtained by reversing the columns of this table.
Note that if y = 5 ^ x, then log{base 5}(y) must be equal to x. For example, 25 = 5^2, so log{base 5}(25) = 2, as you can check from the tables.
The graphs of y = 5 ^ x and y = log{base 5}(x) are shown below.
On the graph a reflection of a point of the y = 5^x graph through the line y = x is a point of the y = log{base 5} (x) graph, and vice versa.

Video File #01

http://youtu.be/oi268Y9RVDs

The figure below summarizes some of the most important properties of logarithms in terms of the base-5 log and its inverse function 5^2.

The first property has already been discussed.
Just below it is the second property, which states that 5 ^ (log{base 5}(x)) = x.

This is merely a statement of the fact that the function y = log{base 5}(x) is inverse to the function y = 5^x.

The other part of this property states that log{base 5}(5^x) = x; this is just another statement of the inverse function property, wherein one function 'undoes' the other.

We can compare these inverse properties to those of the x^2 and `sqrt(x) functions, as shown below.

Another very important and useful property of logarithms is illustrated by the fact that log{base 5}(x) = log(x) / log(5).
This is important because our calculators have a log key but they do not have a log{base 5} key. So we have to use this property to evaluate a base five log.

An example of how this last property might be used is the calculation of log {base 5}(17).

We can use a calculator to calculate this asked log(17) / log(5); we will get a result around 1.7 (the careful not to divide the 17 by the 5 before taking the logs).
If we take 5^1.7, we obtain a number close to 17; if we use the more precise value of the logarithm, we will get a result even closer to 17.
Note that the same result can be obtained by taking natural logs, calculating ln(17) / ln(5).

Video File #02

http://youtu.be/wB_8H3PxTHc

Everything said about the log to the base 5 function applies to any other base, with the 5 replaced by the appropriate base.

We can apply what we now know about logarithms to the task of solving the equation 7 ^ (3x) = 12, as below.

The problem we have run into in previous attempts to solve this sort of equation is that x is stuck in the exponent and we had no way to get it out.
We now have a function inverse to the 7^x function, namely the log{base 7} function.
If we take the base-7 log of both sides of the equation, as in the second step below, we can apply the inverse function property and rewrite the left-hand side has just 3x.
The variable is no longer stuck in the exponent, so we can proceed to solve the equation.
In the fourth line we leave the left-hand side alone while we rewrite the right-hand side log{base 7} (12) as (log 12) / (log 7), which is approximately 1.28.
We can then proceed to divide both sides by 3 in order to obtain x = .437 (approx.).

We can use logarithms to find the exact doubling time tDoub of any exponentially increasing function.

For example, to find the doubling time of y = $10,000 (1.07) ^ t, recall that we must solve the equation $10,000 (1.07) ^ (t + tDoub) = 2($10,000 (1.07) ^ t).
Following the usual steps (see previous class notes), recall that we obtain 1.07 ^ tDoub = 2.
We can easily solve this equation by taking the base-1.07 logarithm of both sides.
Using the inverse function property we obtain tDoub = log{base 1.07} (2), which we easily calculate as log (2)/log (1.07) = 10.2.

Video File #03

http://youtu.be/8wRUWe-QI2M

The number e and the natural function ln(x)

Recall that the number e can be thought of as the limiting yearly growth ratio if we are permitted to compound 100 percent interest continuously.

The figure below shows that if we compound interest at the rate r = 100% = 1 one time, we obtain growth factor (1 + 1/1) ^ 1 = 2 and double our money in a year.
If, however, we take r = 50% = .5 interest twice in a year, we obtain growth factor (1 + 1/2), which we apply twice to obtain a yearly growth ratio of (1 + 1/2) ^ 2 = 2.25.

Applying the same idea, we could take 1/10 of our 100% interest 10 times, obtaining a growth factor of (1 + 1/10), which we apply 10 times to get a yearly growth ratio of 2.59.

If we break our interest rate into 100 parts and apply it 100 times, we obtain approximate growth ratio 2.71. You should verify this.

If we break our interest into smaller and smaller pieces, applying it a correspondingly greater and greater number of times, we obtain the limiting number which we call e, and which has approximate value 2.718281828 (not repeating as it appears to the irrational with an infinite number of decimal places and no repeating pattern).

This number e is the factor by which your principal would grow in a year at 100 percent, compounded continuously.

If interest is compounded continuously at rate r, rather than at 100 percent, we obtain the exponential principal function P(t) = P0 e^(rt).

We note in red in the figure below that e^(rt) = (e^r) ^ t, so are function could be written as P(t) = P0 (e^r) ^ t.
From this form we see that the growth factor for this exponential function is e^r.

Just as we can construct the function log{base b}(t) as the function inverse to y = b^t, we can construct function log{base e}(t) inverse to function y = e^t.

This function, log{base e}(t), is called the natural logarithm function and is generally designated by ln(t).

The ln(x) function is inverse to the e^x function on your calculator.

Video File #04

http://youtu.be/Pi_o3XYCGxU

Properties of Logarithms

Logarithms have a number of important properties, many of which we have already seen.

For example, we have already seen through the example of the base 5 logarithm how log{base b}(b^x) = x by the inverse function property.

Since logarithm functions are inverse to exponential functions, their properties are closely related to those of exponential functions.

For example, we know that x^a * x^b = x^(a+b), which tells us that when we multiply to different powers of the same base we can simply raise that base to the sum of the powers.

So in this sense multiplication can be associated with addition.

We can say the same thing about logarithms. The corresponding law of logarithms tells us that if we take the logarithm of the product of two numbers, the result is a same as if we take the sum of the log of those numbers: log(a*b) = log(a) + log(b).

Similarly, since (x ^ a) ^ b = x ^ (a * b), it can be shown that log (a ^ b) = b * log(a); in both the exponential law and the logarithm law exponentiation is related to multiplication.

Video File #05

http://youtu.be/iCH6D8eCLMg

The first three laws are the most fundamental of the laws of logarithms, in the sense that all the other laws can be derived from these three.

The remaining four laws are, however, useful enough that we include them in our list.
Two of these laws we have seen.

The law for log (a/b) can be derived as log (a/b) = log(a b^-1) = log(a) + log(b^-1) = log(a) + (-1) log(b) = log(a) - log(b).

The last law can be obtained by inspecting the table for any logarithm.

For example, the table for the 5^x function has the point (0,1), since 5^0 = 1.

The table of the base-5 logarithm must therefore contain the point (1,0), indicating that log {base 5}(1) = 0.

It should be clear that the same could be said of any base, since for any number b, b^0 = 1.

Linearizing data

Suppose that we have the data in the table to the left in the figure below.

We notice that y is not a linear function of x, since the y values changed by increasing amounts as x changes by the same amount.

This is too bad because a straight-line fit to data is always a cause for great joy.

Using as a pretty broad hint the fact that all the y values are perfect squares, we attempt to linearize our data by taking the square roots of all the y values.

We therefore say that we use the function `sqrt(y) to attempt to linearize the data.

We obtain the table in the center of the figure, which we can see is linear, with a rise of 2 for every run of 1.

A graph of our transformed data can now be fit with a straight line.
Then, as usual, we use two points on the line as a basis for our linear-function model.
If this is done carefully, we will find that the function y = 2 x fits our linearized data.

All this is great, but what we really want is a model of our original data.

Fortunately it is easy to reverse the linearization.

We merely apply the inverse of the function we used to linearize the data to our linear function model.

Our model was y = 2x, we linearized our data by taking the square root of all the y values.

We therefore reverse the linearization by taking the square of all our y values.

This changes our linearized y = 2x to our final model y = (2x) ^ 2 = 4 x^2.

We check our y = 4 x ^ 2 function against the original table.

If x = 1, this function gives us y = 4, in agreement with the table.
If x = 3, the function y = 4 x ^ 2 gives us y = 36, also in agreement with the table.
The function agrees with the table for the other values of x also, as you should verify.

This example illustrates the process of

linearizing a set of data,
obtaining a linear function model for the linearized data, and then
reversing the linearization by applying the inverse function of the linear function model.

Video File #06

http://youtu.be/d9TKAIGxKMw