But I'm going to get back to it very soon. I just need to fix the website now... get MathJax working and all that good stuff! We'll see! Watch this space. 

Ah, inequality. That terrible, terrible thing. What does one do when confronted with inequality? Well, usually we march, protest, or legislate against it. What do mathematicians do? We work out its domain and range.

So what's stumping me tonight? Well, many things are stumping me tonight, but I feel like I'm getting into my groove. What in particular is stumping me however is this:

So I multiply both sides by the denominator:

Which gives me:

That, my friends, is not a true statement. So what did I do wrong?

More hardcore, actual mathematics will be coming soon. But that requires me to actually have MathJax (the natively installed mathtype software that I've installed on the server) to be up and running and functioning within the blogging software. And then I have to go back and convert all the image based math equations into MathJax typeset. In the mean time - at my last study group, we began talking about symbology. In our last class, we learned that the overbar not only means a repeating decimal, but also means the conjugate of something. Which got me thinking about the below. I don't suggest you try this on your math teacher until I've actually taken over the Mathematical world and imposed my own definitions upon it, but still - take a think about how things you wouldn't normally think of doing can be applied in different ways.

When we see the following equation:

We immediately think - oh! That's a perfect square! Or, at least, maths teachers would desperately hope that's what we think of. And so we factor it:

Now say we see the following equation.

Again, maths teachers would hope we jump up and shout "that's a difference of squares!" And so, we factor it.

But there's no shortcut! But, hang on.

Look! It's an overbar! And an overbar, in this context, could be defined (by me, of course) to mean "take this expression, and multiply it by its conjugate."

So, if we accept the definition of Mathematics by Simon (MbS for short) what would this multiply out to?

A cursory glance of the internet tells me that no-one uses the bar over the exponent to mean something else, but if you know better, please tell me!

It's actually killing me. 

If you haven't already, check out the post below this to see the problem I was having. Now, back to me.

My mistake was elementary. With the second term, when distributing the t into the other two, I distributed incorrectly. So we started with this: 

Let's take a look at that second term more closely. First time around I distributed like this: 

But what I should have done was this: 

Let's look at it another way and commute the multiplication in this term. I'll finish the problem after the break. 

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Currently in the process of doing prerequisite review for my pre-calculus 7A-7B. Either the review material is more difficult than the stuff I learned in MathPath1, or I've forgotten more than I'd care to admit. Either way, the current problem that has me stumped is as follows:

So, looking at this, my first thought is to multiply both sides by the LCD.

This leaves me with the expression:

Bits divide out nicely, leaving me with:

This multiples out and simplifies to:

And that doesn't factor nicely. Now, I could go to completing the square, but according to the computer, this whole problem is meant to be equivalent to a first-degree linear equation. So, people - what have I missed?

Incidentally, I hate fractions.

Hopefully I'll get the mathtype situation sorted soon, because in less than a fortnight I will begin my college-level precalculus MathPath programme! I will be taking Math 7A and Math 7B (Mathematical Analysis, college algebra and trigonometry) in one semester. I'll also be taking a speech and film class, so there's a lot that I'll be taking. 

I'm a little nervous - this is probably the first material I haven't ever seen. I saw some trigonometry during British high school, but it was quite rudimentary, and a cursory look at the textbook feels rather overwhelming. Then again, so did a lot of the material in 131 before I actually dug into and understood, so, we'll see! 

Keep watching this space, updates will come fast and furious once the term starts! 

And yes, the answer to the superman problem is coming. My problem has been trying to implement the technology behind MathJax, in order to properly display equations and such. The method I was using before (an automatic image generation server) seems to have failed, as you can tell by the lack of equations in my blog. 

Good news is I got an A on my 131 course! Pre-calculus algebra and trigonometry, here I come! I'll be back as soon as the technology is working again. 

Been studying for my Math 131 final. It's kind of hellish, especially since our professor is insisting on teaching us new material right up until we take the final exam, which we take the very last day of the semester. We're currently working on inequalities. But, that's why I've been quiet. 

More of a post will follow but in the mean time, look at the below word problem, and see if you can figure it out. I'll be posting the answer after my final. 

"I have 400 grams of Kryptonite. Kryptonite decays continuously at a rate of 4.5%. Fortunately, it only takes 50 grams to kill Superman. But I need a lot of time to plan my fiendish plot. How much time do I have before the Kryptonite reaches the minimum dose? 

Hint: The mathematic constant e measures continuous growth or decay at exponential rates. So, you want a formula that looks like this: where K is the amount of Kryptonite, t is the time, and r is the rate of decay.

My last exam was on functions. I was pretty happy with it, except for a problem which read like this: 

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. . 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. 

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