Precalculus I Class 03/25


During the preceding class we saw that the power function y = 5 x^3 can be linearized with a log(y) vs. log(x) graph, which has slope 3 and y-intercept .699.

We will develop the ideas of logarithms and by the end of class we will see how to solve the equation log(y) = 3 log(x) + .699 for y.

 

We observe that y = log(x) is the inverse of y = 10^x, as can be verified using a calculator.

You should check these claims out using your calculator. 

The figure below shows these relationships schematically.  We note the following:

 

The base b logarithm function is the inverse of the b^x function, as indicated in the table below:

Properties of exponents are related to properties of logarithms, as indicated in the table below.

We can illustrate the law that says that log{base b}(x) = log(x) / log(b) by looking at the relationship between the base-e logarithm ln(x) and the base-10 logarithm log(x):

Since ln(x) = log{base e}(x), which by our law is equal to log(x) / log(e), our law tells us that ln(x) should be equal to log(x) / log(e).

This tells us that the function y = ln(x) is a vertical stretch by factor 2.3 of y = log(x).

We now use the laws of logarithms to solve the equation log(y) = 3 log(x) + .699.

We begin with the equation log(y) = 3 log(x) + .699, and we raise 10 to a power equal to each side of the equation to obtain the equation

10^(log y) = 10 ^ (3 log x + .699). 

Since 10^(log y) = y by the inverse-function relationship we obtain

y = 10 ^ (3 log x + .699).

Starting with y = 10 ^ (3 log x + .699) we complete the solution:

Since b^(x+y) = b^x b^y (where x and y are not related to the x and y in the equation) we have 10 ^ ( 3 log x + .699) = 10^(3 log x) * 10^.699 and the equation becomes

y = 10^(3 log x) * 10^.699.

10^.699 is just a number we can evaluate on our calculator.  10^(3 log x) = (10^(log(x) ) ^3, and 10^log(x) = x so (10^(log(x) ) ^3 = x^3 and we now have

y = x ^ 3 * 10^.699.

Evaluating 10^.699 we obtain 5 so our final equation is

y = 5 x^3.

Note that this is the equation we linearized in the previous class to obtain the equation we just solved.  So we've come full circle.  The process was:

In this example we knew the function y = 5 x^3 to start with.  Had we not known that the data follows from this equation, we would have found the equation anyway.