## Logarithms Explained, and the Associative Property of Multiplication

So much advanced math relies on a firm grasp of basic Algebra and Algebra II.

Today, lets take a look at logarithms!

So what are logarithms? Well, first let’s look at exponential equations, such as $2^x = y$ where the 2 is a base. We all know that for example, $2^3 = 8$. A general form is $b^x = y$ where b is the base. Well, with logarithms, the format is $log_b y = x$. So for $2^3 = 8$, we would express that with logarithms as $log_2 8=3$. Fun, isn’t it! The logarithm is the number that the base is raised to a power by to equal a given number; in the example above, the base 2 is raised by the power 3 to equal the number 8.

So the tricky part is that you get rules like $log_b y + log_b z = log_b (yz)$ Huh? How’s that? Well, it has to do with the Associative Property of Multiplication and the rules for multiplying exponents. Basically, when you multiply exponents with the same base, such as $b^m * b^n$, you add the two powers; so the answer is $b^{m+n}$. So for example, $2^2 * 2^3 = 2^{2+3} = 2^5 = 25$ . Keep in mind that the Associate Rule of Multiplication says that (ab)c=a(bc); so $2^2 = (2*2)$, $2^3 = (2*2*2)$, and $2^2 * 2^3 = (2*2) * (2*2*2)=(2*2*2*2*2)=2^5$. These rules are important for working with logarithms.

So, why is the following true: $log_b y + log_b z = log_b (yz)$? What it’s saying is if you add what you
raise b by to get y plus what you raise b by to get z, you’ll get what you raise b by to get yz. Now, imagine an expression $log_2 8 + log_2 16$. Why would the equation $log_b y + log_b z = log_b (yz)$ work here? Well, $log_2 8=3$ and $log_2 16=4$. By $log_b y + log_b z = log_b (yz)$, the left hand side, $log_2 8 + log_2 16=7$, should equal the right hand side $log_b(yz)$ . Here $y=8$ and $z=16$, so $yz=128$. If you work out the math, $log_2 8 + log_2 16=7$, and with these numbers, $log_b (yz)$= $log_2 (8*16)=log_2 (128)=7$,

, so indeed $log_2 8 + log_2 16$=$log_2 (8*16)$=$log_2 (128)=7$. Wow, so $log_b y + log_b z = log_b (yz)$ works! Why? A way of looking at this is, $log_2 8=3$ is equivalent to $2^3 = 8$, and
$log_2 16=4$ is equivalent to $2^4=16$ . Since $y=8$ and $2^3= 8$, and $z=16$ and $2^4=16$ , $yz$= $2^3 * 2^4$ = $8*16$ which by the rule for multiplying exponents = $2^7=128$. And $2^7= log_2 (128)$. And viewing this by the associative rule, $2^3= 2*2*2$ and $2^4 = 2*2*2*2$ , so $2^3$ *$2^4= 8*16$= $(2*2*2)*(2*2*2*2)$ = $2^7 =128$. Then
$log_2 8 + log_2 16= log_2(2* 2* 2) + log_2(2* 2* 2* 2) =$ $log_2 (2* 2* 2* 2* 2* 2* 2 )= log_2 128$,
the general form of which is $log_b y + log_b z = log_b (yz)$. I think.

Logarithms are great because you can solve exponential equations with them. You can do this with this log rule: $log_b (m^n) = n * log_b m$. So for example, if you had the equation $2^x = 50$, you could start taking the log of both sides, $log (2^x) = log 50$ , which equals $x* log 2 = log 50$, which equals $x= log 50/log 2$. Then you can just figure out log 50 and log 2 with a calculator, divide, and you’ve got your answer. You sometimes might want to use the change of base formula too. It’s important to remember that $b^{log_b m}= m$, this will come in handy when solving some equations. For example if you had the equation $x^2 = 50$, you could take the log of both sides, $log (x^2) = log 50$, which equals $2* log (x) = log 50$, which equals $log (x) = log 50/2$. Now, how to isolate the x? Here we’re using base 10. So raise en to the power of each side of the equation, $10^{log (x)} = 10^{log 50/2}$. That equals $x = 10^{log 50/2}$, you can solve the right hand side with a calculator to find out log 50 and divide that by 2, and raise ten to the power of the result. Basically when you have $log_b x = y$ what that means is $b^y = x$. So if $log_b x = y$, if you raise the base b to y which equals $log_b x$, you will get simply x.

**These equations were designed in Microsoft Equation Editor (comes free with Microsoft Word)–but then to my chagrin I realized I don’t know if or how I can export the document with the equations to a format such as text plus image files where the formatting of equations (subscripts, superscripts) would be preserved in a blog post. Until I get scribd/ipaper working or figure out how they put those equations in Wikipedia (do people insert the jpgs of the equations one by one by hand, or do they use a program that can convert the text and equations into text and images all at once, and can Microsoft Word do this? Here’s the original PDF file; I still want to be able to use iPaper for ease, I then figured out how to render the equations in Latex for WordPress, but it seems kind of manually laborious? Is there some easy way to do this all? Here’s some more info on using Latex with WordPress.com