My last exam was on functions. I was pretty happy with it, except for a problem which read like this:
=x^2+2x+1\\ \text{Solve:} f^{-1}(f(0)) $)
Easy enough on paper - an function composed with its inverse leaves your plugged in value behind. In this case, the function of 0 inverted is just 0. BUT. f(x) doesn't have an inverse function, cause it's a quadratic. I'm hoping it wasn't a trick question by the professor, and more hoping that we'd recognize that functions are undone by their inverses. The rest of the exam I felt reasonably alright about.
I feel a lot happier today about logarithms than I did a few days ago. I could see they were blue hippos, but I couldn't see whether I could multiply two blue hippos to get a green one. One of my main sticking points, though, is one of the main points of logarithms and their operations - properties of logarithms. Much like properties of exponents! Well, that's cause they are exponents.
Exponential functions and logarithmic functions are going to be my next exam. Two more exams, then a final. And then, assuming I pass this section of algebra, I'll move on to Math 7A and 7B, pre-calc! Fun times, huh. So. For the people here for math.
This is an exponential equation. 
. You can say that as "two to the power of two equals four."
And this is the logarithmic equivalent of that same equation. =2 $)
. That's pronounced "the log of 4 base 2 equals 2." In other words, the number that two gets raised to in order to get four is equal to two. Logarithms are exponents. The argument (the bit in parentheses) is what you end up with when you raise the base (the big subscripted next to the log) to a power (the logarithm itself.) So the logarithm of 27 when the base is 3 is
3, because three to the third power equals 27.
=3 $)
The two expressions above aren't equal. Rather, they're inverses of each other. It works something like this:
=y=a^x\\f^{-1}(x)=log_a(y)=x $)
The logarithmic equation is the inverse of an exponential equation.
So. Properties of logarithms. I'm not going to go into too much detail, but for useful revision for myself and my class (and anyone else wanting to know about the properties of logs) I present them below in easy list f
ormat:
=log_b(A)+log_b(C) $)
This allows you to take the product of two logs and split them up separately. For example:
=log_2(32)=log_2(4)+log_2(8)\\4\cdot 8=2^2 \cdot 2^3=2^{2+3}=2^5\\32=2^5 $)
They're related in exactly the same way. Just like you add exponents when multiplying two numbers with the same bases but different powers, so too can you add the logarithms of two numbers with the same bases to get the right exponent. =log_b(A)-log_b(C) $)
This is exactly like the one above, but instead of using products, you use quotients. =C\cdot log_b(A)+log_b(C) $)
Now this one is VERY useful, especially when doing a change of base on an exponential equation. For example, if I wanted to find X below algebraically:
I can solve for x by taking the logarithm of both sides and solving for x like so:
=log(32)\\x log(2)=log(32)\\x=\frac{log(32)}{log(2)}\\x=5 $)
There's more to why this particular function works. I'll explain more in a later post, but for now, if you remember that in one to one functions (which logarithms in most cases are), you can prove the 1-1 relationship by saying if a=b, then f(a)=f(b). By the same token, if a=b, then log(a)=log(b). And logarithms are just numbers, so you can divide, multiply, add, and subtract them just like any other variable or number when dealing with equations.
Next post will be on change of base formula. Also coming up look for a brief post on proving that a function is 1-1.
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