College Algebra: Expanding Logarithmic Expressions

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Now let’s try to see these new formulas that we just came up with in action, by trying to do the following. Someone will give us a logarithmic expression and I want to, basically, simplify it and just write it as a sum or differences of logarithms as best I can using these properties. So, let’s recap the properties. Remember, that logb b is just 1. Log b of 1 is just 0. Those you can just think about and see by the definitions of logarithms. The logarithms of a product is the sum of the logs. The logarithms of a quotient is the difference of the logs and a logarithms of xy power, that y can be pulled out in front as a coefficient--and I see y logb x.
So, let’s see if we can see those things in action with some of these things. So, remember that the question in each case is, “Can you simplify this logarithmic expression to be just the sum or differences of logs and nothing else?” Log3 of 3a divided by b. Well, here’s how I think about it. I see this as a log of a quotient. The log of a quotient is the differences of the logs. So, using that formula, I can write this as log3 of 3a - log 3 of b. These two things are equal by that formula. Log of a quotient is the difference in logs. What about the 3a? Well, that’s actually a product. So, the log of a product is the sum of logs. So, this equals log3 of 3 plus log3 of a, and then I still have this term, minus log3 of b. But, what about log3 of 3? Well, that’s something we should know. What number is that? That’s the exponent I have to raise 3 to in order to make it equal 3.
What exponent of 3 gives me 3? Just one. So, this can be simplified even further to 1 + log3 A - log3 B. So, there is this expression in the equivalent form, but just using sums and differences of logs, in a very simple way. I was able to untangle all that product and quotient thing into this formulation. These are equal by the properties of logarithms. Okay? That’s why I used the formulas. Let’s try some more. So much fun.
Log6 of 6 square root of 5 divided by square root of 6. Well, here again I see a quotient. So, I could write this as log6 of the top, 6 square root of 5, minus log6 of the bottom, square root of 6. But what else can I do. Well, this is actually the log of a product. So, that’s the sum of the logs. So, this equals log6 6 + log6 of the square root of 5, minus log6 of the square root of 6. Maybe that’s the best that we can do. Times that, actually, we can make this a little bit simpler. First of all, log6 of 6, that’s just 1 again. These square roots are actually ½ exponents. Let me write that in. This is 5 to the ½ exponent - log6 of 6 to the ½ exponent. The reason why that’s useful is because now you can remember that property, that if I have inside the log something to a power, that power can come out as a coefficient. That was that last property that we looked at.
Let me show you that one again. That was this property right here. See, log of something to a power--I can pull that power out in front as a coefficient. If I use that right now, I can actually simplify this a little more. This is 1 +, and now that ½ is out in front, log6 of 5 minus, and now that ½ is out in front, log6 of 6. Oh wait a minute, log6 of 6 is 1 again. So, this number is just -½ and I’ve got a 1 out in front, so 1 - ½ is just ½. So, I see this equals ½ + ½ log6 5. That is the simplification of this really complicated thing. This complicated thing is identical in numerical value to ½ + ½ log6 5. I was able to find that by just simplifying.
So, you can see how dramatically easy it is. If you to evaluate this thing, it could be hard. But evaluating this was pretty easy. Just find log6 5, using a calculator or something, multiply it by ½ and then add ½ to it. You get it, instead of doing all that complicated stuff. Just again illustrating the power of these properties of logarithms. Let’s try another one.
Log3 (A + 3B). Well, some may say, “Okay, the log of a sum is the sum of the logs. So, this is log3 of A + 3log3 of 3B.” But they would be wrong. This is classic mistake number 9. The log mistake. The sum of the logs is not the log of the sum. So, there is no simple way to write this. This is just itself. There’s no way to make this any simpler. There’s no formula for that. Even though you might say, “Hey, wait, the product of the of the product here is the sum of the logs.” Yes, but I can’t undo that. So, that’s the best you can do. A little trick problem there to sort of get you thinking.
All right, one last one. Log10 the fifth root of A²B4 all divided by C³. Looks really awful but watch how we can do this. First of all, a fifth root is the same thing as raising something to the 1/5 power. So, this is just log10 of A²B4 over C³ and I raise the whole thing to the 1/5 power. Well, now that 1/5 by one of these laws of logarithms--that 1/5, log of something to 1/5, I can bring that out in front. So, when I do that, I see that this equals 1/5 in front times log10 of A²B4 over C³. Well, that can be simplified a lot. I see a log of a quotient. That’s the difference of the logs, using that property that we talked about earlier. So, what I see here is the following.
What I see is that this equals the difference--I still have that 1/5 out in front, log10 of A²B4. Then I have a minus--and I actually have that 1/5 in front of there, or I could put a big parenthesis around it if you want it, log C³. So, there’s the log of a quotient, difference of the logs. What does that equal? I still have that 1/5 way out in front of everybody. What do I see now? Now, I see--well, this is log of a product. So, that’s the sum of the logs. I just keep using these formulas again and again to really drill this home. So, log10 of A² + log10 of B4 and then, don’t forget, I’ve got that minus log of C³. Now, I see log of something squared, log of something to the fourth, log of something cubed. So, I can take each of those exponents and bring it out in front. This is a really great property, if you’re careful with it.
1/5 times 2log10A+ 4log10B - 3log10C. You can distribute that 1/5 if you want it throughout. I’ll do that. I don’t know why. So, I have 2/5, here, log10A + 4/5log10B - 3/5log10C. You’ll remember the original question was log10 of the fifth root of all that stuff. I was able to rewrite that into this very tidy form. Well, not tidy, but it’s very simple. It’s just log terms that I’m adding or subtracting together with coefficients. So, what you can see is using these properties of logs, we’re actually able to express these kind of objects in a much simpler way, just by doing this. So, that’s sort of neat. By the way, one little teeny, stupid thing. Don’t ever read this as log 10 times A. It’s not. It’s a function, log10 of A or log10 evaluated at A. Just a little teeny point. Log always has to have something that you’re taking the log of. I cant say, square, because you’ve gonna say, “Well, square of what?” I can’t square root, you’ve got to say square root of 25 and then you know 5.
Anyway, there are some wonderful examples where you can just simplify complicated looking things and make them look--well, many pieces, but each piece easier, just using the properties of logs. That’s the point, property of logs. Enjoy.