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October 08, 2025, 10:27:45 pm

Author Topic: VCE Methods Question Thread!  (Read 5723142 times)  Share 

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shadows

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Re: VCE Methods Question Thread!
« Reply #3420 on: December 29, 2013, 12:52:19 pm »
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Can anyone explain what this is used for?
logb (N) = loga (N) / loga (b)

Thanks  :)

although a lot of schools don't cover it and theres hardly any questions on it... do learn it!
There was a mtlp choice q in a past vcaa exam where you need to know the change of base formula.

alchemy

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Re: VCE Methods Question Thread!
« Reply #3421 on: December 29, 2013, 01:28:39 pm »
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I've got a conflicting idea regarding the following question  :-\

Question: 2x=11. Solve for x without using a calculator.
I rearranged this to log211=x. This is as far as I got.
However, looking at the worked examples for similar questions it asks to take log10 off (divide by) both sides of the original equation. Doing this I get:
log102x=log1011
x=log1011/log102.

Notice how both expressions how the latter expression for x conflicts with the initial. What is the correct method of solving this problem?

Zealous

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Re: VCE Methods Question Thread!
« Reply #3422 on: December 29, 2013, 01:37:56 pm »
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There was a mtlp choice q in a past vcaa exam where you need to know the change of base formula.
I believe is was VCAA 2011 MCQ21 from memory.
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Conic

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Re: VCE Methods Question Thread!
« Reply #3423 on: December 29, 2013, 01:38:26 pm »
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I've got a conflicting idea regarding the following question  :-\

Question: 2x=11. Solve for x without using a calculator.
I rearranged this to log211=x. This is as far as I got.
However, looking at the worked examples for similar questions it asks to take log10 off (divide by) both sides of the original equation. Doing this I get:
log102x=log1011
x=log1011/log102.

Notice how both expressions how the latter expression for x conflicts with the initial. What is the correct method of solving this problem?

log1011/log102 is the same as log211. I'd usually go with log211.
« Last Edit: December 29, 2013, 01:41:26 pm by Conic »
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Re: VCE Methods Question Thread!
« Reply #3424 on: December 29, 2013, 01:42:33 pm »
+1
If you are asked to solve it without a calculator, then you normally turn it into a log of base 10. Back before CAS calculators, to find a numerical approximation for , you'd convert it to a log of base 10 and then go look up a log table for the value.

I'll just note that the latter expression doesn't conflict with the former, they are actually equivalent using the change of base of log formula stated earlier in this thread.


Which means , which as can be any positive number, we choose as we don't have a calculator to do it for you (then in the old days you'd go look up the log table, but that's not required anymore).

Both give the same value, but in the context of this question the second method would be the one you should go by.
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BLACKCATT

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Re: VCE Methods Question Thread!
« Reply #3425 on: December 29, 2013, 01:54:10 pm »
-1
I believe is was VCAA 2011 MCQ21 from memory.

dat memory

alchemy

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Re: VCE Methods Question Thread!
« Reply #3426 on: December 29, 2013, 03:17:26 pm »
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Both give the same value, but in the context of this question the second method would be the one you should go by.

So only the expression for x should be specified in the answer? The worked examples give the answer in decimals, but I'm assuming the worked examples were done using a calculator.

b^3

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Re: VCE Methods Question Thread!
« Reply #3427 on: December 29, 2013, 03:39:01 pm »
+1
So only the expression for x should be specified in the answer? The worked examples give the answer in decimals, but I'm assuming the worked examples were done using a calculator.
I'm guessing you're using the Essentials Textbook right? It's a good textbook (probably the better one for methods anyways), but here is a case of a question that is a little bit redundant. They've left in that question to teach students how to use the change of base rule, but since you aren't given log tables anymore, the last bit is done on the calculator (even though it says not to..., because you'd need a log table that you don't have and don't need for the course anymore...). If anything, it's something that wasn't quite changed properly when CAS was introduced into the course.

Now, what you should take from this is to know how to use the change of base rule when you need to apply it, that's all you need to know for these types of questions.

The whole decimals without a calculator seemed a bit odd to me when I first did that exercise a few years ago, but after realising what they were trying to do, it makes more sense (I didn't know about log tables and such during year 12, because you don't need to know about them :P).

tl;dr, The question isn't great, just know how to use the change of base formula.
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Bluegirl

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Re: VCE Methods Question Thread!
« Reply #3428 on: December 30, 2013, 12:19:22 pm »
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Could someone help explain how to get the answer. I know you factorise but I'm lost.

x^3 - 2x^2 - 7x - 4=0

I'm really stuggling at the moment and I'm losing confidence :(

« Last Edit: December 30, 2013, 12:23:58 pm by Bluegirl »

abcdqd

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Re: VCE Methods Question Thread!
« Reply #3429 on: December 30, 2013, 12:47:47 pm »
+2
Could someone help explain how to get the answer. I know you factorise but I'm lost.

x^3 - 2x^2 - 7x - 4=0

I'm really stuggling at the moment and I'm losing confidence :(
To factorise a cubic, we have to first find a linear factor, then use polynomial long division to divide the cubic by this factor, which gives us a quadratic factor.
Recall the factor theorem: If , then is a factor of .
So what value of , when subbed in to the equation, will make it equal 0? Try subbing in 1, -1, 2,.. We find that makes the equation equal to 0.
Therefore, is a factor. Now perform long division, .
Answer
We should get
So
Factorising the second term

so
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alchemy

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Re: VCE Methods Question Thread!
« Reply #3430 on: December 31, 2013, 09:23:01 am »
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Polynomials.  Anything i should know about as i'm starting it.

Just make sure you learn them in the right order. Quadratics first, cubics second, quartics third, ...
Also make sure you know what a polynomial is and isn't before you start off. Makes life easier later on.

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Re: VCE Methods Question Thread!
« Reply #3431 on: January 01, 2014, 03:15:19 pm »
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Hi can anybody help me with this question:

Find two quadratic functions f and g such that f(1)= 0, g(1)=0 and f(0)=10, g(0)=10 and both have a maximum value of 18

Thanks

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Re: VCE Methods Question Thread!
« Reply #3432 on: January 01, 2014, 03:42:15 pm »
+2
For the quadratic to have a maximum, we need to have a parabola where the coefficient on the term is negative. Since we know the maximum value will occur at the turning point, we can use the turning point form for a parabola, , but will need  a negative out the front to give us a maximum, so . Our turning point will be at , so since our maximum is , . So we have , . Now for and we have two pieces of information each, and we have two unknowns each. So we can substitute these points in and solve for the two unknowns by solving the two simultaneous equations formed. You'll get two pairs of solutions corresponding to the two different equations.

Have a go first before looking here
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psyxwar

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Re: VCE Methods Question Thread!
« Reply #3433 on: January 01, 2014, 04:02:45 pm »
+3
Hi can anybody help me with this question:

Find two quadratic functions f and g such that f(1)= 0, g(1)=0 and f(0)=10, g(0)=10 and both have a maximum value of 18

Thanks
therefore
therefore , rearranging we get
Thus,

The graph has a maximum value at , or substituting our back in, at



Solving this for a (typing in LaTeX is too hard...), we get and

Using this we can work out the value of b () and thus f(x) (and g(x), which just uses the other set of a and b values we didn't use).

and

edit: b^3's method looks easier to do tech-free, go with that
« Last Edit: January 01, 2014, 04:05:01 pm by psyxwar »
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RedCapsicum

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Re: VCE Methods Question Thread!
« Reply #3434 on: January 01, 2014, 04:21:08 pm »
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Cheers for the solutions!

Yeah, I was able to figure out how to do the problem by hand by b^3's explanation but thanks for the additional working out :)