Hi, and welcome. My name is Anthony Varela. And today we're going to apply the properties of logs. So we will review these logarithmic properties and then use them to simplify and evaluate logarithmic expressions. So as a reminder, log base b of y equals x is the logarithmic equation. Equivalently, we have the exponential equation b raised to the power of x equals y.
And here are our important logarithmic properties. We have the product property of logs, which allows us to take a product inside of a log, so x times y, and we're going to break this down, then, into the log base b of x plus the log base b of y. We have our quotient property, which is very similar to the product property. This just has division inside of a log instead of multiplication, and we have the log base b of x minus the log base b of y, instead of addition, with our product property.
We have a power property to logarithms, where if we have an exponent power inside of a log function, we can bring that outside of the log as a scalar multiplier. So log base 2 b of x to the n equals n times log base b of x.
Now, we also have our change of base formula, which is particularly useful for using your calculator to evaluate logs. More than likely, your calculator will have a log button under base 10 and a log button under base e. So what would we do, then, if our base of the log was neither 10 nor e? Well, the log base b of x can be written as the log base a of x over the log base a of b. So for example, the log base 7 of x can be rewritten as the common log of x over the common log of 7. That's how we would use change of base.
And then we have the log base b of 1 equals 0, and the log base b of b equals 1.
So we're going to be using these logarithmic properties to simplify and evaluate log expressions. Our first example is blog base 2 of 8 plus log base 2 of 32 plus log base 2 of 2. So one of the things that I noticed right away is that each of my log expressions is a base 2, and they're being added together. So I can make use of our product property. We're going from something that looks like this into something that looks like this. I'm going to multiply 8 by 32 by 2. That's 512. So this is a single log base 2 of 512.
Well, I can interpret log base 2 of 512 as being 2 raised to some power and equaling 512. So x, then, would be the value of log base 2 of 512. Well, 2 to the 9th power equals 512, so x equals 9.
Now, another way I can think about, then, evaluating the log base 2 of 512 is I can rewrite 512 as 2 to the 9th power. And I can use my power property to bring 9 out in front. So I have 9 times the log base 2 of 2. And here, the argument to the log is equivalent to the base. So this evaluates to 1. So we see, then, that this expression equals 9.
Another way to think about evaluating this expression is to evaluate each log individually, and then add them together. So thinking about the log base 2 of 8. Well, this would be 2 raised to some power equals 8. 2 cubed equals 8. So log base 2 of 8 equals 3. Log base 2 of 32 would mean that 2 raised to some power equals 32. And 2 raised to the 5th power is 32. So log base 2 of 32 is 5. And log base 2 of 2-- this can be written as 2 raised to some power equals 2. And we saw from before, that that equals 1. So 3 plus 5 plus 1 is 9. Same number that we got before.
Now, here in our next example, we have the common log of 2x over y. And we are given that the log of x equals 2.5 and the log of y equals 1.8. Now be careful here. We cannot substitute 2.5 in for x and 1.8 in for y, because we're given that the log of x is 2.5 and the log of y is 1.8. So I need to first simplify this expression. So I see that I have a quotient inside of the log function. So I'm going to write this as the log of our numerator 2x minus the log of our denominator y. Well, now I have a substitution that I can make. I know that log of y is 1.8. So this is the log of 2x minus 1.8.
Now, inside of this log function, I have a product 2 times x. So I'm going to use my product property to break the log of 2x into the log of 2 plus the log of x. And we have another substitution we can make, then, with a log of x. So log of 2, I'm going to approximate as 0.3, and then I'm going to add the log of x, which we know is 2.5, and take away the log of y, which we know is 1.8. So our expression then evaluates to 1.
Here, we have a final example of log base 5 of x cubed over 2. And we're given that the log base 10 of x equals 1.08. Remember, if there's not a base explicitly written, it's a base 10. And what's going to be important here is that this is written with a base of 5. And we need to express that in the log base 10 of x in order to substitute that with 1.08.
So how are we going to simplify this expression? Well, I have a quotient inside of this log function. So let's use our quotient property. This would be log base 5 of x cubed minus log base 5 of 2. Now, inside of this log function, I have an exponent. So I can use my power property to bring an exponent out in front. So we have 3 times log base 5 of x minus log base 5 of 2.
Now I can't make my substitution quite yet, because this is a log base 5 of x. So I need to use change of base to rewrite so that I have log base 10 of x in order to substitute. So how I'm going to do that, well, I know that I'm going to have log of 5 in my denominator, because that's the base to these two logs here. So this would be 3 times common log of x minus common log of 2. So now I can substitute log of x with 1.08. So 3 times log x is 3 times 1.08, so 3.24. I'm going to subtract the log of 2, which is approximately 0.3, and then divide by the log of 5, which is approximately 0.7. So then evaluating 3.24 minus 0.3 divided by 0.7, our expression, then, is approximately 4.2.
So let's review our lesson on applying the properties of logs. Well, we have properties involving products, quotients, and powers. And we also have our change of base, which allows us to rewrite the log expression in a different base so that our calculator might be able to evaluate it for us. And we also have the log base b of 1 equals 0. And the log base b of b equals 1. Thanks for watching this tutorial on applying the properties of logs hope to see you next time.