Now that you have an idea of what a logarithm is, let's explore their properties. Since the logarithm of a number is an exponent. Let's actually review the properties of exponents first, so that way, we can see how they apply to logarithms. What's the result of each of these operations? You should write your answer using x as the base, and then using some combination of a and b for the exponent.
Here are the three expressions. Great work if you got each of these correct. When we multiply bases, we add these exponents together. So x to the a times x to the b equals x to the a plus b. And when we raise a power to another power, we really multiply these exponents. So to the a raised to the power equals x to the a times b. And finally, when we divide the same base, we subtract the exponents. We have x to the a minus b. Now, keep in mind that the logarithm of a number is an exponent. So, we know that these properties are going to have some sort of relationship with logarithms. Let's look at some examples and figure it out.
The first property states that if we add 2 logarithm's together that have the same base then we can just multiply the numbers together. For example if we have log base 2 of 4 plus log base 2 of 8 then we know that equals log base 2 of 4 times times, which equals 32. Now, this statement should make sense. We know that log 2 of 4 equals 2. And we know that log base two of 8 equals three. So if we add those together, we should get 5. And sure enough, the log base 2 of 32 equals 5. So whenever you add logarithms that have the same base, you can really just multiply these arguments together. We just multiply x times y and then take the log of it. So let's see if you can extend this reasoning. If we have log base b of j, plus log base b of a, plus log base b of m, what would that equal, if we did the log base b of this expression? In other words what expression goes right here so that way the right-hand side of our equation is equal to the left-hand side of our equation.
The left hand side is equal to the log base b of j times a times m. Remember when we add two logs with the same base, we can really just multiply these two numbers together. So we can simplify these first two logarithms to be the log of base b of j times a. Now we apply this rule one more time, and we multiply j times a by m. So we'll just have one logarithm, log of base b, of j times a times m. This is how we know this left-hand side is equal to this right-hand side. I use quotes here just to represent the same expression written here and here.
The next property of logarithms has to do with exponents, or powers. When we have log base b of x to the a, we can really set that equal to a times log base b of x. In other words, we just take this exponent of a and we bring it down in front of our logarithm as a coefficient. In other words, if I had log of base down in front of our logarithm, so we'll have 3 times the logarithm of base 10, of 100. Now I usually put this in parentheses, so I understand that it's 3 times this quantity. And let's carry out the map on both sides of the equation to make sure that this makes sense. On the left hand side we'll have log of base 10 and then 100 to the 3rd power. Well 100 to the 3rd power is the same thing as a million. That's 1 followed by 6 zeroes. On the left hand side we'll have 3 times the log base ten of 100. Log base 10 of 100 is 2. We know this, since 10 squared equals 100. We're looking for the exponent that we'd raise 10 to, in order to get this number. So, we know 3 times 2 equals 6. And now let's check the left-hand side. We have log base 10 of a million. But we know a million is 1 followed by 6 zeros. So, that means we have to raise 10 to the 6th power in order to get a million. So yes, 6 equals 6. This is just one example of why this log property holds true. And finally, here's the last logarithm property we'll examine. If we subtract two logarithms that have the same base, then we can just divide the two numbers. So the log with base b of x minus the log of base b of y equals the log of base b of x divided by y. For example, the log of base 3 of 81, minus the log of base 3 of 9 would equal the log of base 3 of 81 divided by 9. We simply just divide these 2 numbers inside of our argument and take the log of it. And let's also be sure that this checks. We know the log of base 3 of 81 equals 4 since 3 to the 4th equals 81. We also know the log of base 3 of 9 equals 2, since 3 squared equals 9. And, finally, on the right-hand side, we'll have the log of base 3 of 81 divided by 9. We know 81 divided by 9 really equals 9, so we have log of base 3 of 9. Now we can just subtract here, so we get 2 equal to 2. We know log of base 3 of 9 is really just 2, as before. Now, using those log properties, I want you to try and write this expression with 1 logarithm. In other words, you should have the log with base b of 1 number here. Now it's okay if you're not entirely sure what to do, I just want to try your best. Think back to what it means when you add logs together and when you subtract logs together. Do them one at a time and I think you might get the answer. And again it's totally okay if you don't get there. You can always see the solution.
This expression is equal to the log of base b of 5. Great thinking if you got this answer. Now, to start, I think it's best to think about this as two logarithms of base b of 5. If I write this line all out. I see I have log base b of 2 plus log base b of 5. Plus log base b of 5, minus log base b of 10. And notice that there are two log base b's of 5 here. Now I'm ready to apply my properties of logarithms. I know when I add two logarithms together that have the same base, I can really just multiply these numbers together. So have log base b of two times five. I'm going to do this one more time. I have one more addition here, so I can multiply these number times these numbers. So we have log base b of 2 times 5 times 5, minus log base b of 10. Now when we subtract logarithms, we take this number and divide it by this number. So we'll have log base b of 2 times 5 times 5 divided by 10. Now, you might be wondering why I didn't go ahead and multiply these numbers. And it's because I knew that I would have to simplify in the end. So I can see that 2 times 5 times equals 10, and that reduces with another ten in the denominator. These simplify to one. And this is how we get our final result of log base b of 5.
If that first problem got you thinking, then try this one. Apply these logarithm properties again, and then write your final answer here as one logarithm.
It turns out that this answer is also log base b of 5. Great work if you got this right. Again, I think it's best to think about splitting this up into two logarithms. Log base b of 5 and log base b of 5. We're adding three logarithms together right here. So, we can really just multiply these arguments together. So, we'll have log base b of 3 times 5 times 5. That's how we can combine these three logarithms through addition. And then we'll have minus log base b of 15 on the end. And finally to subtract two logarithms with the same base, we'll divide the second one into the first one. So we'll have log base b of three times five times five divided by 15. We have the factors of 15, which reduced to one, which leaves us with log base b of five. Now, there was one other way we could have solved this problem and it involves using the power rule. Let's see what that would look like. We could've started by moving this 2, backup as an exponent for the 5. So 2 times log base be a 5 would equal log based b a 5 squared. Now we can apply the same rules for adding logger items with the same base. And subtracting logarithms with the same base. When we add these 2 logarithms together, we'll have log base b of 3 times 5 squared. We simply multiply these two numbers together. And finally here, we'll divide this number into this number. So to combine these logarithms when subtracting, we'll have the log base b of 3 times 5 squared divided by 15. Notice that 5 squared is the same thing as 5 times 5. So we're really back to what we ended with last time. Our factor of 15 simplifies to 1 so we're left with log base b of 5.
How about this third problem. Can you write this as a single logarithm?
This expression is equal to log base b of 6. Great work if you found this logarithm. Now again, I think it's easiest to think about having two logarithms of base b of 8, and three logarithms of base b of 4. We know that to combine these three logarithms, or to add them together, we'll need to multiply these numbers together. So, we can express the addition of these three logarithms as the logarithm of base b of 6 times 8 times 8. Then we'll have the subtraction of log base b of 4, three times. When we subtract a logarithm, we really divide this number into this number. Keep in mind this is only true if the logarithms have the same base. So when I combine this logarithm with this logarithm, we'll really just divide 6 times 8 times 8, by 4, and we'll take the log of base b of that expression. These other two logarithms are still on the end. I notice that I'm going to have to repeat this process. So, if I take this logarithm with base b and subtract this logarithm with base b, I'm going to divide by another end. I need to subtract this logarithm of base b of 4 from this logarithm of base b of this expression. So, I'm simply dividing by another 4. This gives me a logarithm with one number. Now, I just need to simplify. I know 8 times 8 equals 64; and 4 times 4 times 4 also equals 64. So, these factors simplify to
Try expressing this expression as a single logarithm. Be sure that your logarithm uses the base of a and uses the variables of r, s, and t. Good luck.
This expression can be written as the log of base a of r times t cubed divided by s. Excellent work if you found this log rhythem. We can start by doing the division of r and s since we have log of a minus another log of a. We simply divide s into r and combine the logarithms together. Now we're going to use the power rule to bring this coefficient of 3 up as the exponent for t. So, 3 times the log of base a of t is equal to log of base a of t cubed. We're adding two logarithms together that have the same base. So we can just multiply r divided by s, times t cubed, here. We know t cubed is the same thing as t cubed divided by 1. So when I multiply these fractions together, I just multiply their numerators to get r times t cubed. And we multiply the denominators s and 1 to get s. Here's our single logarithm.
How about this expression? How do you think you could write this using a log of base 10 with 1 argument?
This expression equals the log of base 10 of x squared minus 16. Here's one logarithm and one expression. Now that's great reasoning if you figured this one out. You really just want to multiply the x plus four times the x minus 4. Since we have two logarithms added together with the same base, we just multiply these two arguments together. So we'll have the log of base 10 of this entire product. This is a difference of square pattern. A plus B times A minus B, so we'll know when we multiple we'll get A squared minus B squared. For x squared minus 16. We also want to be very careful. This might be confusing, because we're not sure if we're taking the log of base 10 of x squared, or the log of base 10 of this entire quantity. That's why it's best to use parentheses when you want to indicate that. Here I would think I am taking the log of base want to take the log of base 10 of this entire number. Whatever x squared minus answers. Be sure you know what you're taking the log of.
As a challenge problem to you, try and write this expression using only one logarithm. Your answer should be in the form log of base b of some expression. Whatever this expression is, should involve the variables x, y, the number 5 and the other variable t. Now, there won't be a solution here since this is a challenge question, but I hope you figure it out.