How Logarithms and Exponents are Related ● Logarithms and Exponents are inverse functions of each other.o Logarithms are the exponent that is necessary to raise a number to produce a given number. o Exponents are the logarithm of a number whose base is 1.● The product rule is used to calculate the derivative of the product of two functions. The quotient rule is used to calculate the derivative of the quotient of two functions. The powerand root rule is used to calculate the derivative of a power or root function. o The product rule states that if f and g are differentiable, then so is f × g. That is, we can find the derivative of their product by multiplying their derivatives. o The quotient rule states that if f and g are differentiable, then so is f ÷ g. That is, we can find the derivative of their quotient by dividing their derivatives. o The power and root rule states that if x is a real number, then ex has a power function with base e raised to x as its exponent: ex = e^x When you understand the concept of exponents, then understanding logarithms should not bemuch more difficult. The reason is that the two concepts are closely related; a logarithm is reallythe answer to the question raised to what power? For example, if you multiply 2 by 2 by 2 or 8 andthen want to know the power being used, the answer is 2 to the 3rd. Same story with other logarithms. The logarithm to the base 2 of 8 is 3. The logarithm to the base 2of 16 would be 4. The logarithm to the base 2 of 32 would be 5. Like that. In the next couple of videos, we'll talk about logarithms. I hope that after you work some examples,you'll see that they conform to some simple rules and they are less scary than they might seem atfirst. A formula for a logarithm can be written in two different forms. Its exponential form can be expressed as a number raised to the power x, resulting in anothernumber. To use the example I just gave, you can set b=2, x=3, and 2 to the power of x. Here, we write log tothe base b. We use the letter b because we are referring to the base of a logarithm of N. So, log to the base 2(8)=3, and this is a logarithmic interpretation of the same idea: 2 to the 3=8.So, 2 to the 4=16, where log2(16)=4. What is b? b is called the base. Thus, that's 2. What is N? Nis 16. The exponent X is 4. The exponential form and the logarithmic form express the same idea, which is the relationshipbetween three numbers: a base, an exponent, and the result. So, as you may remember from high school mathematics, any number to the power 0 is equal to 1.Similarly, the log of any base is equal to 0 when that base is 1. Since 2 to the 0 is equal to 1, 10raised to the 0 is equal to 1, and 20 to the 0 is equal to 1 (when expressed in exponential form), afew basic rules can be used to solve problems involving logarithms. Now three of the most important rules to know are the product rule, the quotient rule, and thepower and root rule. The product rule for logarithms looks like this: log of a product equals log of X plus log of Y. In thisformula, I don't have to specify the base as long as I keep it the same throughout the problem. Itworks for any base. So, any log(xy) = log x + log y, where x and y are variables in the equation.Now let's do an example. The logarithm of 35, or log b(35), is equal to the sum of the logarithms of7 and 5: log b(7) + log b(5). The same way, the quotient rule states that if I have the log of a fraction or a quotient, X over Y, thisis equal to log X minus log Y. If I had log2 64, this would be the same as log2 64 minus log2 2.
The Power and Root Rule is used when we want to take the logarithm of a number raised to apower. This is equivalent to nlogx. In certain situations, this approach makes it much easier to solve otherwise difficult problems. As an example, we can take the logarithm of the square root of a number, which is simply one-halfthe logarithm of that number. This rule works for positive or negative integers and fractions. So weknow that 35=7 times five. Thus, this equation is equal to logb(7) + logb(5). 35 is equal to 70/2.Thus, it would also be equal to logb(70)-logb(2). The logarithm of 16 divided by 4 is equal to thelogarithm of 16 minus the logarithm of 4. The logarithm of 16 is 2 to what power, or base 4? Thatwould be 2 x 2 x 2 x 2, or just 2 to the power of 4. The logarithm of 4 to the base 2 is 2. Solog2(16/4) is equal to 4 minus 2, which equals 2. Here are other illustrations. Cube root of 1,000 can be expressed as one-third log2(1,000). To go from log10(7) to the fifthpower, you simply multiply log10(7) by itself five times. Log of anything of x to the -1 should bewritten as - logb of x. Well, that's pretty much it. Those are our basic rules. And if you apply them, you can simplify justabout any logarithm. Let's try applying several rules at once, shall we? We are about to use our product and root rule, but product rule comes first. We see this equalslogb of x squared + logb of y to the -3, which of course is equal to 2 times the log base b of x minus3 times the log base b of y. Now it’s time for the quotient rule. Thus, the logarithm of x2, divided by x to the minus 1⁄2, is equalto the logarithm of x2 minus the logarithm of x to the minus 1⁄2. It means we have 2logb of x. Now we have a minus, a minus that has been reinterpreted as plus one-half log b of y. Here's a useful hint. If you have log a + log b = log a × b, it's also true that log a - log 1/b = log a ×b. So we just put in a minus 1 and a minus 1. Logarithms are also useful for solving problems in which exponents are involved. For example, wecan treat both sides of an equation as if they contained an exponent. We can treat both sides of alogarithmic equation, or both sides of an exponential equation, as if they were exponents of thesame number. Let me use an example to demonstrate. We will now calculate log2(39x/x-5)=4. To do so, we will take 2 to the log of all of this. 2 to thefourth power = 2 to the log2 of that thing. So we got 39x/x-5=16. We can then multiply both sides ofthis equation by x-5, and simplify. So we have 39x=16x and 80. Now, we subtract 16x from both sides. So, we have 23x = -80. We divide both sides by 23 to obtainx = -80/23.