Once you have a basic understanding of exponents, it should not be too much of a leap to understand logarithms. The reason why is that these two concepts are very closely related. A logarithm is really the answer to the question raised to what power? So for example, if you have 2 times 2 times 2 or 8, and you want to know what power is being use, well, it's 2 to the 3rd. In just the same way, the logarithm to the base 2 of 8 is 3. The logarithm to the base 2 of 16 would be 4. The logarithm to the base 2 of 32 would be 5 and so on. So, we'll be talking about logarithms the next couple videos. And I hope that after you've worked some examples, you'll see that just like exponents they follow some simple basic rules and they're really not as intimidating as they might have seemed. So, a formula for logarithms can have two different forms. It can have the exponential form where we have a number raised to the power x and this equals something. So, to use the example that I just used, you can have b=2, x= 3 and 2 to the 3=8, which is N. Or, we can express the same idea in logarithmic form. And here, we write log to the base b. We use the letter b, because we are referring to the base of a logarithm of the number N. So log to the base 2(8)=X=3, and that's the logarithmic form of the same idea, namely that 2 to the 3=8. So 2 to the 4=16, log2(16)=4. What is b? b is the base. So that's 2. What is N? N is 16. And what is the exponent X? That would be 4. Okay, so we have the exponential form and the logarithmic form, and they capture the same idea which is the relationship between three numbers, a base, an exponent and the result. So, just as we have the simple rule when raising any number to the power 0, any number to the power 0, you may remember is equal to 1. Similarly, the log to any base, this number here is the base, of 1 is equal to 0. Right, why? Because 2 to the 0=1. 10 to the 0=1. 20 to the 0=1, when expressed in exponential form. Just as with exponents, we have a few basic rules that we will use to solve and simplify problems that contain logarithms. Our three primary rules are the product rule, the quotient rule and the power and root rule. The product rule looks like this. Log of a product (XY). Now when I write log without a base, what this notation means is that this is true for any base. So, this is a general formula and I don't need to specify the base as long as I keep the base the same within any given problem. So any log(XY) = log X + log Y. So, let's do a quick example. We have logb(35). This is equal to log b(7) + log b(5), because 7 times 5 equals 35. Very similarly, the quotient rule says that if I have the log of a fraction or a quotient, X over Y, this is equal to log X- log Y. So if I had log of 64 divided by 2, this would be the same as log of 64 minus log of 2. Power and Root Rule is the rule that we apply when we have a number raised to a power, and we want to take its logarithm. This is equal to nlogx. So there are many situations when this makes it quite easy to solve an otherwise hard problems. So for example, we might have the log of the square root of some number and this is simply going to be one-half the log of that number. This rules works when N is positive or negative, an integer or a fraction. So we know that 35=7 times five. So this will be equal to logb(7) + logb(5). 35 is also equal to 70/2. So this would also be equal to logb(70)- logb(2). Log2(16/4). Well, that should be obvious that, that would be equal to log2(16)- log2(4). Log2(16) is 2 raised to what power equals 16? So, that would be 2 times 2 times 2 times 2, or 2 to the 4. Log2(4) would be 2 raised to what power equals 4? So that would be 2 times 2, or 2 to the 2, okay? So log2(16/4) is going to be equal to 4 minus 2 or 2. Okay, let's look at a few more examples. Log2(1,000), cube root of 1,000, would simply be one-third log2(1,000) Log10(7) to the fifth would simply be 5 times log10(7). Log of anything of x to the -1, would be- logb of x. So, those are our basic rules. And if you apply them, you can simplify just about any logarithm. So let's try combining several rules together at the same time, okay? We're going to use our product and root rule, but first we're going to use our product rule. So we know that this is equal to logb of x squared + logb of y to the -3, which of course is equal to 2logb of x- 3logb of y, okay? Now, we apply our quotient rule. So, logb of x squared divided by x to the minus one-half is going to be equal to logb of x squared minus logb of x to the minus one-half. So, we have 2logb of x. And now we have a minus, a minus which is a plus. + one-half logb of y, okay?. Here's a handy trick just in general. When we have log a + log b = log of a times b. It's also going to b equal to log a- log of 1 over b. So we simply put in a minus 1 and a minus 1. And there are problems where that is very useful. One of the sophisticated ways that we can use logarithms is to treat both sides of a logarithmic equation as if we had an exponent. So we have an equation, either a logarithmic or an exponential equation of the form x=y, and we treat both sides of that equation as if they were exponents of the same number. Let me show you what I mean. Okay, we have log2(39x/x-5)=4. So, what we're going to do is we're going to just take 2 to the log2 of all of this. = 2 to the 4. So, 2 to the log2 of anything simply equals that thing. So we have 39x/x-5=16. Then we can multiply both sides of this equation by x-5, and we can simplify. So we have 39x=16x. 80. Now, we subtract 16x from both sides. So, we have 23x = -80. We have x = -80/23.