Unit
Inverse, Exponential and Logarithmic Functions
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
To unlock all 5,300 videos, start your free trial.
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Expanding logarithms. When we expand we want to be able to write a complicated logarithm as a series of more simple logarithms and in order to do this what we have to do is use the properties of logs. Here I am looking at a quotient; one thing divided by another, so I know that I can use my quotient rule logarithms and split them up as subtraction.
We have log base 9 of quantity (ab)³ minus log base 9 of 9 (c plus d). There’s a couple ways we tackle it from here. We’re dealing with the log of ab to the third. Now this third is associated with everything inside of this log, so what we could do is bring this 3 around to the front. That’s one option.
The other thing we could do is distribute this 3 in, so we get a³b³, that would be another option. Doesn’t matter which you do as long as you make sure that this 3 is somehow associated with both the a and the b. I’m going to take it out in front but you could distribute it through if you wanted that as well.
So this is going to be 3 log base 9 of a times b. I’m just going to work through this one by itself and then we’ll go tackle this one afterwards. So a times b, we have our product inside of our logarithms so we could split this up. Log 9 of a times b is going to be split up as log base 9 of a plus log base 9 of b, but what you need to know is that this term is the same as that. We still need to bring this 3 down and put it out in front because the 3 is going to be associated with both the a and the b so we end up with 3 log 9 of a, plus 3 log 9 of b.
We have this component. We also need to do a similar expansion on this term over here. We are multiplying so when we multiply we could split it up as addition, so this turns into log base 9 of 9 plus log base 9 of c plus d. But again similar to this 3 over here, where I distributed through this is actually a minus sign, so we have to minus this entire thing. We can either distribute it now or later it doesn’t matter, as long as at some point it gets distributed through.
Using our rules of logs, log base 9 of 9 is just 1, so this is going to be minus 1 and then plus whenever we are adding inside of a log, remember we can’t do anything with that. If we're multiplying we could split it up, if we’re dividing we could split it up but if we’re adding or subtracting this is stuck to be the same thing. This is the log base 9 of c plus d and lastly making sure we then distribute that negative sign through, minus 1 minus log base 9 of c plus d.
Fairly long involved answer but what we have done is using the rules of logarithms we have expanded a fairly complicated logarithm into the product, the sum and coefficients of a number of more simple logs.