Unit
Exponential and Logarithmic Functions
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
To unlock all 5,300 videos, start your free trial.
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
I'm going to talk about the change of base theorem. Change of base theorem is a great theorem because it allows you to change logarithms from one base to another. I want to start with an example, express log base 5 of 20 in terms of base 10 logs.
The first thing I want to do is, set up an equation, y equals log base 5 of 20. I'm basically just naming the log base 5 of 20y and then I can take the definition of logs to write this in exponential form. So this will be 5 to the y equals 20.
Now since I want to change to base 10 logs, I'm going to take a log of both sides, the base 10 logs of both sides. So log of 5 to the y equals log of 2. And now log over power property, I can pull the y outside, y times log of 5 equals log of 20. And finally, since I want to find the value of y, I divide both sides by log 5. So dividing by log 5, I get log of 20 divided by log 5 and that's it. Log base 5 of 20 is log of 20 over log of 5.
Now this suggests the following theorem. If want to change from log base a of x to any other base say of base b, it's going to be log base b of x over log base b of a. Now the way I remember this is, first of all I'm changing from base a to base b, which one of these goes on top? Well x is higher than the a, so x goes on top and a goes on the bottom, that's the way I remember it.
Let's use this in an example. This problem says evaluate each by switching to another base. So let's switch this to base 2. So we have a log base 2 of 16 over the log base 2 of 8. The log base 2 of 16 is 4 and log base 2 of 8 is 3, so I get 4/3.
Here, log base 16 of 8, this is the log base 2 of 8 over the log base 2 of 16, that's 3 over 4. This sort of interesting phenomenon here, if you notice I've switched the base and the argument and I got the reciprocal answer. I'm going to come back to that in a second.
In this example, both of these numbers are very nice powers of 10, so I'm going to switch to the log base 10, the common log. So this will be log 0.00001 over the log of the cube root of 10. Now the log of .00001, this is 10 to the -5, so the log of 10 to the -5 is -5. The cube root of 10 is 10 to the 1/3, so log of 10 to the 1/3 is 1/3. -5 over 1/3 is the same as -5 times 3 which is -15.
I said I was going to come back to this point right here. When you change the base and the argument of log, look at what happens to the output, you get the reciprocal. To state that as a theorem, log base a of b is 1 over the log base b of a. This is actually a corollary of the change of base theorem. And so to use that in an example, log base 81 of 3, I really would like the situation reversed, the log base 3 of 81 is much nicer. So by this corollary, log base 81 of 3 is 1 over the log base 3 of 81 and I know that 81 is 3 to the 4th power, so this is 1 over 4.