VCE Methods Question Thread!

[Skitty]

If there is a function,
y = 2^(x+1) - 2
and a second function y = 2^x
[

• ,[y]]=[
• ,[y]]+[[c],[d]]

How would you find the values of c and d that map of y = 2^x to y = 2^(x+1) - 2
Thanks

I recommend you check this out and try to apply it to your question. https://mathspace.co/learn/world-of-maths/functions-graphs-and-behaviour/matrix-transformations-of-functions-52385/matrix-transformations-of-functions-1904/

[vcestressed]

Can someone please help me with this question?
solve for x:
4^(x+2) = 27^(1-3x)

[Skitty]

Can someone please help me with this question?
solve for x:
4^(x+2) = 27^(1-3x)

4^(x+2)=27^(1-3x)
2^(2x+4)=3^(3-9x)
log2(2^2x+4)=log2(3^(3-9x))
2x+4=(3-9x)log2(3)
2x+4=3log2(3)-9log2(3)x
2x+9log2(3)x=3log2(3)-4
x=[3log2(3)-4]/[2+9log2(3)]

So basically change base, expand and collect like terms.

Hope this helped ^^

[vcestressed]

4^(x+2)=27^(1-3x)
2^(2x+4)=3^(3-9x)
log2(2^2x+4)=log2(3^(3-9x))
2x+4=(3-9x)log2(3)
2x+4=3log2(3)-9log2(3)x
2x+9log2(3)x=3log2(3)-4
x=[3log2(3)-4]/[2+9log2(3)]

So basically change base, expand and collect like terms.

Hope this helped ^^

Hey the ans is actually x=log27-log16/log4+log27?

[Shadowxo]

Hey the ans is actually x=log27-log16/log4+log27?

There are often a few ways to express things, especially with logs.

Skitty's answer is just in a different form. You could also use different bases.
Also, when writing maths expressions, brackets are always helpful (your answer wasn't very clear)

[vcestressed]

Makes sense now. Thanks Skitty and Shadowxo

[Skitty]

Hey the ans is actually x=log27-log16/log4+log27?

Was trying to simplify it further sorry if you confused you

There are often a few ways to express things, especially with logs.

Skitty's answer is just in a different form. You could also use different bases.
Also, when writing maths expressions, brackets are always helpful (your answer wasn't very clear)

Still dunno how to format it like that

[lzxnl]

Was trying to simplify it further sorry if you confused you

Still dunno how to format it like that

I haven't posted on these forums in a while.

I'll rewrite what Shadowxo posted twice. Once in LaTex form, and once without so that you can see the syntax. Then you can go play with it yourself. That's how I learned LaTex myself. Also, this isn't quite the same LaTex that you use for documents; there are very small differences that I won't go into.



Slight differences here because I like using the cross instead of asterisk
And because having brackets for logs is clearer.
Here's the source code. You'll need the two tex's I had above at the start and end of the code to make the computer output actual maths symbols.

4^x \times 4^2 = 27^1 \times 27^{-3x}\\
16\times 4^x = \frac{27}{27^3 x}\\
4^x\times 27^{3x} = \frac{27}{16}\\
(4\times 27^3)^x = \frac{27}{16}\\
x = \frac{\ln\left(\frac{27}{16}\right)}{\ln(4\times 27^3)}\\
= \frac{\ln(27) - \ln(16)}{\ln(4)+\ln(27^3)}\\
= \frac{\ln(27)-\ln(16)}{\ln(4)+3\ln(27)}

[Skitty]

I haven't posted on these forums in a while.

I'll rewrite what Shadowxo posted twice. Once in LaTex form, and once without so that you can see the syntax. Then you can go play with it yourself. That's how I learned LaTex myself. Also, this isn't quite the same LaTex that you use for documents; there are very small differences that I won't go into.



Slight differences here because I like using the cross instead of asterisk
And because having brackets for logs is clearer.
Here's the source code. You'll need the two tex's I had above at the start and end of the code to make the computer output actual maths symbols.

4^x \times 4^2 = 27^1 \times 27^{-3x}\\
16\times 4^x = \frac{27}{27^3 x}\\
4^x\times 27^{3x} = \frac{27}{16}\\
(4\times 27^3)^x = \frac{27}{16}\\
x = \frac{\ln\left(\frac{27}{16}\right)}{\ln(4\times 27^3)}\\
= \frac{\ln(27) - \ln(16)}{\ln(4)+\ln(27^3)}\\
= \frac{\ln(27)-\ln(16)}{\ln(4)+3\ln(27)}

Thanks dude. How do you manage to do the question like that? Seems pretty hard to work out what you're doing whilst typing in this latex format, or is  it something that'll be natural with practise?

Also good to see a 99.95 dude that isn't going mainstream. I'm considering maths as well.

[lzxnl]

Thanks dude. How do you manage to do the question like that? Seems pretty hard to work out what you're doing whilst typing in this latex format, or is  it something that'll be natural with practise?

Also good to see a 99.95 dude that isn't going mainstream. I'm considering maths as well.

:')
Honestly, a high ATAR really shouldn't make you do something you didn't originally want to. The point of an ATAR is to get into what you want. Who cares if it's not med? I probably would have dropped out of medicine if I went in this year after BSc because I cannot stand rote learning.

It really is something that comes with practice. I routinely type up assignments and my revision notes now using LaTex, so I can visualise if I'm missing brackets or where I'm currently typing. However, it's also a good idea to regularly preview your work too to see.

[Shadowxo]

Was trying to simplify it further sorry if you confused you

Still dunno how to format it like that

There's a great Latex Guide by RuiAce which will help if you want to know how to format it
Also, if someone's written in Latex you can use the "quote" feature to see what they've written

Thanks dude. How do you manage to do the question like that? Seems pretty hard to work out what you're doing whilst typing in this latex format, or is  it something that'll be natural with practise?

Also good to see a 99.95 dude that isn't going mainstream. I'm considering maths as well.

Latex becomes easier with practice, once you get the hang of it gets easy to figure out what to do

[Skitty]

Thanks for the replies! I might start practising Latex, looks pretty neat.

:')
Honestly, a high ATAR really shouldn't make you do something you didn't originally want to. The point of an ATAR is to get into what you want. Who cares if it's not med? I probably would have dropped out of medicine if I went in this year after BSc because I cannot stand rote learning.

It really is something that comes with practice. I routinely type up assignments and my revision notes now using LaTex, so I can visualise if I'm missing brackets or where I'm currently typing. However, it's also a good idea to regularly preview your work too to see.

Yeh I agree nowadays people are pressured into doing courses which require higher scores rather than what they actually prefer. I used to think like this in middle school ( I told everyone imma do Med), but thankfully I've come to realise that it's about what you want to do and what you enjoy.

[lzxnl]

Yeh I agree nowadays people are pressured into doing courses which require higher scores rather than what they actually prefer. I used to think like this in middle school ( I told everyone imma do Med), but thankfully I've come to realise that it's about what you want to do and what you enjoy.

Indeed. Don't underestimate how much hell the wrong course could bring upon you. Not saying you won't enjoy med, not saying you'll enjoy maths, but doing medicine if you're not set on it is a risk you need to be prepared to take.

[VanillaRice]

Sorry for the bad quality (and that its upside down XD) but could someone please explain to me how they have calculated the value of 'a' (TAKEN FROM MARK SCHEME)

I also get a=3 but I get this by solving for a in 6=a(4-3)^3 + 3 BUT I'm concerned that maybe I dont understand a better method (e.g the one in the mark scheme

Also, shouldn't 'h' technically be 3 as the equation is provided in the form f(x)=a(x MINUS h)^3 +k

Thanks in advance 

You've used the same method as the answers - except you've substituted (4,6), while the suggested answers have substituted (2,0). Most likely reason for this is that it's easier to work with zeros

Also, I agree with you, I think h should be = 3.

Hope this helps

[Mattjbr2]

Can someone please explain to me the transformations which lead you from what I circled in green to what I circled in red?
I.e. how to go from: \[(x+1)^2 -1\]
to \[16((x-\frac{1}{2})^2 -\frac{1}{16})\]
I put them both into "standard" form to make the transformations more clear. But I'm confused as to how my answer is incorrect.
My answer is:  dilate 16 units away from the x axis, followed by a translation of 3/2 to the right and 15/16 up.
The book's answer is:  dilate by a factor of 1/4 from the y-axis, then translation 3/4 units to the right
I considered using the matrix transformation thing but I'm not sure

This is from Cambridge Chapter 8, Q9b page 334