VCE Methods Question Thread!

[brightsky]

Why is it a factor of 5 though? Is it because its then it becomes 5? And does it really matter, in terms of losing marks if i continue to list transformations as i already do?

Something like that. Dilating a graph by a factor of 5 from the x-axis has the effect of replacing the variable t with the variable t/5. Note that it is always the variable DIVIDED by the dilation factor, never multiplied.

Consider the graph of y = f(x), where f is some random function.

- A dilation by a factor of k from the x-axis has the effect of replacing the variable y with y/k. This yields the equation y/k = f(x), which is equivalent to y = k*f(x). The reason why a dilation from the x-axis affects the variable y rather than the variable x is clear if you consider what happens to a single point when you apply a dilation from the x-axis. The x-coordinate stays the same. It is the y-coordinate that changes.
- A dilation by a factor of k from the y-axis has the effect of replacing the variable x with x/k. This yields the equation y = f(x/k).

Note that no matter what the dilation, it always has the affect of replacing a variable with the variable DIVIDED by the dilation factor.

With regard to wording, I would recommend that you switch to the wording that I used above. It would be incredibly mean of them to deduct marks due to variation in wording, but as they say, better safe than sorry.

[keltingmeith]

For the graph of    When we are solving for x, let y=0 but why cant we do this:









Why doesnt this method work? Its frustrating me so much because it follows the log laws so it should work, right?

Can you elaborate on that? Because it seems to work fine for me.

[keltingmeith]

Now solve with CAS or manually, however why doesnt this work? And hows it fine to you??

Think about this: ln(x) is a one-to-one function, so it should only intercept the x and y axis once. However, you've somehow gotten a quadratic, which can have up to 2 solutions. So, when you get to that quadratic, you've added in more solutions. It's up to you to figure out where. It's important to note that when you add in more solutions, you don't lose the one that you want - you just add in ones you don't want. Checking them in the original expression can help you figure out which ones you don't want.

[DSubShell]

Edit: Read Eulers first and try figure on your own! My answer just expands the "solution" to this problem to more general cases.

For the graph of    When we are solving for x, let y=0 but why cant we do this:









Why doesnt this method work? Its frustrating me so much because it follows the log laws so it should work, right?

Haha I remember this frustration. It's tricky to explain this but I'll try.

Firstly, let's refer to the act of turning the coefficient of a log into a power as "Plogging"

The best way to start is to look at it visually. On your calc, graph the following:
and

Notice how although they are the same shape, they are actually different graphs. The squared version has both branches! So while "Plogging" log law does work, it has the potential to expand the domain of the function, leading to more answers.
This is very similar to when you square an equation, and you get dummy answers.

The general rule of thumb is, that if you introduce an even power/indice into your working somewhere, you will most likely get dummy answers (because of the nature of even powers).

However, the even more general rule of thumb is to always check your answers against the original domain. (ie. In the original, what is inside the log can't be negative). Then if any answers don't fit, just cross them out and provide a reason. (I've seen you in the Specialist thread... so this idea is especially important for specialist!)

So long story short: your working is correct, but always make sure to fit answers to the domain.

Also, as a bonus, here is how to avoid the mess of getting two solutions as an extra:

[keltingmeith]

Here's some food for thought, as well (that I only thought about just then when DSubShell mentioned specialist. A+ explanation, btw - I might steal it the next time I'm asked. ):

You've just discovered that you can "add" solutions when you increase the power of an expression. What do you think could happen if you decrease the power of your expression?

[IndefatigableLover]

Quick question guys:

  why cant we have a y-intercept when x=0? Please explain it in a easy way

The maximal domain of that function is from 2/3 to infinity hence x=0 will not exist (this no y-intercept for when x=0 )

[brightsky]

Because the argument of any logarithm has to be strictly greater than 0. If you substitute x = 0 into the equation, you'll get y = 2 + ln(-2), which is, as far as Methods is concerned, undefined.

You can also think of it this way. Since the argument of any logarithm has to be strictly greater than 0, 3x - 2 > 0, which implies x > 2/3. Hence, the domain of the function is (2/3, infinity). The graph of the function has a vertical asymptote at x = 2/3, and the entire graph lies to the right of that asymptote, so the graph never cuts the y-axis.

[brightsky]

Yeah I know that haha, but in general why cant we have any log of a negative value??

Let b = log_a(NON-POSITIVE NUMBER), where a is a real number not equal to 1 but strictly greater than 0. b is not defined because there does not exist any real number b such that a^b equals a non-positive number.

But if we relax the restriction on b and make it so that b can be a complex number, then b = log_a(NON-POSITIVE NUMBER) is defined. Consider the number w = ln(-1). It can be shown that w = pi*i, where i is the imaginary unit and is defined by the equation i^2 = -1. You can read more about the complex logarithm here: http://en.wikipedia.org/wiki/Complex_logarithm.

[keltingmeith]

For   to work out the x value, why cant we do this method:



   (divided 0 by -2)

  and from now cancel our the log e? It works for most of the other equations, but for this one, what power does e have to be in order to obtain a 0?

Except that , not 0.

(*hint* _e will turn that e into a subscript for ya, works just like ^2. Also, \log looks nicer than log. Also works for ln and circular functions. But the last bit is just being pretty particular, the first bit is the more practical so things are easier to read. )

[brightsky]

ln(e^0) does not equal to 1.

From the fourth step onwards:
1 = ln(x+1)
ln(e) = ln(x+1)
e = x + 1
x = e-1

[brightsky]

What? Isnt anything raised to the power of 0 = 1 ?

I agree that e^0 = 1. But ln(e^0) does not equal to 1. I think the confusion arises because of the LaTeX. If you mean ln(0), then I also do not agree that ln(0) equals to 1. ln(0) is undefined, for reasons that we've already discussed. It is ln(e) that equals to 1.

[keltingmeith]

WHERE AM I GOING WRONG, SO PISSED

That's not how you convert to exponential... Couple of things wrong here:

1) You either haven't defined the log base (at which point one assumes it's base 10, but I don't think VCE maths likes you doing this), or you're using the log base as a number. The second is a BIG no-no. Watch how to solve it using the method you're currently using:



And a faster way:



2) This isn't chemistry, you should define your bases.

[brightsky]

omg no, disregard what I said, i totally understand it now,

because is basically which is zero

but is this the right way i keep getting confused lol

: which is equals to 1 ?

Nailed it.

[keltingmeith]

But if we relax the restriction on b and make it so that b can be a complex number, then b = log_a(NON-POSITIVE NUMBER) is defined. Consider the number w = ln(-1). It can be shown that w = pi*i, where i is the imaginary unit and is defined by the equation i^2 = -1. You can read more about the complex logarithm here: http://en.wikipedia.org/wiki/Complex_logarithm.

Between you and lzxnl, next year's methods/spec cohort is going to have the most extended knowledge, hahah. (granted, pretty sure I do the exact same thing half the time~)

[keltingmeith]

Guys, I just did a similar question and DID THE SAME mistake again! I feel so stupid, is there any way to master these mathematical concepts like my fellow masters have? Please fill me in, I mean TWO similar mistakes in a row is definitely saying something...

Blu-tac the log laws to the back of your bathroom door. Namely and