VCE Methods Question Thread!

[darkz]

Thanks for the first solution!
For the second one, where did you get 'e' from (btw what even does that mean)
I was told by my tutor that if you don't write a base number in front of a log, then it's assumed that it's 10, in this case, that's what I did.

Yeh my bad, read it wrong rip

[fun_jirachi]

e is Euler's number (the 2.718281828...). When taking the log in this base it can be referred to as the natural log but whatever, its totally irrelevant
if it just says log, then it should technically assumed to be in base e. maybe your tutor said that because on the calculators many people use the log button uses base 10, while the ln button uses base e. However, in most/all textbooks log is usually going to be base e, making it equivalent to ln.

[TTanner01]

Q1 is just 2. Recognise from your log laws that

So from that, you get that
For Q2, solve it as a quadratic in (logx). When you see a term that's squared, another that's the term by itself, and a constant, usually its a quadratic. If it makes it easier, let logx equal some constant a, then solve for a, substituting logx=a once you've found your solutions for a.

And from there I'm pretty sure you should be able to find the values of x.

Thank you so much!
How do you expand 2(logx)^2 ?
Your ways good but I wanna try another

[MB_]

Thank you so much!
How do you expand 2(logx)^2 ?
Your ways good but I wanna try another

The other method would be substitute logx for u and solve for u like you usually would in a quadratic and then sub your answers for u back into u = logx and solve for x

[fun_jirachi]

I might be wrong on this one, but you can't expand (logx)^2. Also, any other way is pretty damned convoluted (outside of darkdzn's way which is essentially the same thing i did, just took the extra step of letting logx=a, I skipped that step and solved for logx) or doesnt exist.

The other method would be substitute logx for u and solve for u like you usually would in a quadratic and then sub your answers for u back into u = logx and solve for x

yup that's what darkdzn did

[sailinginwater]

Can someone please explain how to do question 2b of the 2010 methods exam 1?

[Lear]

Can someone please explain how to do question 2b of the 2010 methods exam 1?

Please try to show us at least some sort of attempt at the question when asking these sort of questions. It doesn’t help you in future instances if someone just spoon feeds you the answer rather than correcting where you may have struggled or gone wrong.

[sailinginwater]

Please try to show us at least some sort of attempt at the question when asking these sort of questions. It doesn’t help you in future instances if someone just spoon feeds you the answer rather than correcting where you may have struggled or gone wrong.

Loge(1-x) dx = -loge(1-x) with the top terminal as 3 and the bottom terminal as 2, but it doesn’twork since loge(-2) and log3(-1) are undefined. I checked the solutions and it mentioned absolute values but aren’t they taken out of the course?

[Lear]

Loge(1-x) dx = -loge(1-x) with the top terminal as 3 and the bottom terminal as 2, but it doesn’t work since loge(-2) and log3(-1) are undefined. I checked the solutions and it mentioned absolute values but aren’t they taken out of the course?

Yeah absolute values are taken out of the course. That exam is from 2010 when they were in the course. You can either ignore that question or just learn what absolute values are. Either way they will not be tested in the exam. You just need to know (like you do already) that you cannot have a negative log so if you are ever are required (unlikely) to integrate 1/x you should include x>0 as the domain.

[sailinginwater]

Yeah absolute values are taken out of the course. That exam is from 2010 when they were in the course. You can either ignore that question or just learn what absolute values are. Either way they will not be tested in the exam. You just need to know (like you do already) that you cannot have a negative log so if you are ever are required (unlikely) to integrate 1/x you should include x>0 as the domain.

But according to my teacher the above question is still in the current study design, so is there a way to do the question without using absolute values?

[Lear]

But according to my teacher the above question is still in the current study design, so is there a way to do the question without using absolute values?

No there isn’t because by definition the anti derivative of that function must be absolute other wise we are evaluating the natural log of negative numbers. I am certain that if VCAA does decide to test your knowledge of the anti derivative of 1/x they will ensure that it will be something like integrate 1/2x+1 between 4 and 5 to ensure if does not matter if you know absolute or not.

[Opengangs]

Notice that:
\[ \int \frac{1}{1-x}\,dx = \left\{\begin{array}{lr}
        -\ln\left(1-x\right), & \text{for } x < 1\\
        -\ln\left(x-1\right), & \text{for } x > 1 \\
        \end{array}\right.\]

This just comes from the definition of an absolute value. Hopefully, this helps!

[sailinginwater]

Is the change of base rule likely to show up in exam 1 and 2?

[Lear]

No one can tell you whether it is likely as no one knows what it is on the exam. It’s assessable and should be learnt well.

[sailinginwater]

Can someone please explain how the change of base rule works?