Year 13 Maths Help Thread

Announcements Posted on
How helpful is our apprenticeship zone? Have your say with our short survey 02-12-2016
    Offline

    3
    ReputationRep:
    (Original post by kiiten)
    Ohhhh so you do e^(ln x^2) = e^8
    X^2 = e^8
    X = sqrt e^8

    ?? Wait ive gone wrong, sorry my minds just gone blank
    Remember that x^2 = 4 is asking "what numbers can I square to get 4?" And the answer is obviously x=2, but don't forget that x = -2 \Rightarrow x^2 = (-2)^2 = -2 \times -2 = 4 also works.

    In general, if you have x^2 = k then your solutions are x = \pm \sqrt{k}, so can you apply this here?
    Online

    2
    ReputationRep:
    (Original post by RDKGames)
    That is worrying if you're confused on that. Remember that \ln(x^2)=\log_e(x^2), and you SHOULD be able to know what happens here after Core 2.

    Not sure how else to explain it to you. You should know that you cannot have a logarithm of a negative number, therefore \ln(x) cannot have x as negatives. When it comes to \ln(x^2) however, then you CAN have negative x values because negatives/positives squared just become positive.

    When you go from \ln(x^2) to 2\ln(x) you are essentially getting rid off all the negative x-values hence getting rid off one of the solutions. Get it?
    Kinda. Please could you remind me of that core 2 example you were talking about (ive forgotten most of it :3 )

    Ok so if there are no -ve values why is the answet + or - ?? When it cant be a -ve? (I realised im over complicating it - sorry).
    Offline

    3
    ReputationRep:
    Fs.
    It's just x^2=|x|^2


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by kiiten)
    Ok so if there are no -ve values why is the answet + or - ?? When it cant be a -ve? (I realised im over complicating it - sorry).
    When you have \ln x then x can't be negative, since you can't have a negative number inside the logarithm.

    When you have \ln x^2 then x can be negative, since x^2 will still be positive anyway and you can have that inside the logarithm.
    Offline

    3
    ReputationRep:
    (Original post by kiiten)
    Kinda. Please could you remind me of that core 2 example you were talking about (ive forgotten most of it :3 )
    Here's an example from Core 2. Solve: \displaystyle \log_2(x)=3

    This of course reads as "what is the number I get, x, when I raise the base, 2, to the third power?"

    From here you would fully go with cancelling the logarithm by making both sides an index of a certain number, that number being the base of the logarithm:

    \displaystyle 2^{\log_2(x)}=2^3

    and you should know that \displaystyle a^{\log_a(b)}=b therefore the equation becomes:

    \displaystyle x=2^3

    Now you can apply the same idea to your equation without moving the exponent of x down.

    (Original post by kiiten)
    Ok so if there are no -ve values why is the answet + or - ?? When it cant be a -ve? (I realised im over complicating it - sorry).
    No, there ARE negative 'x' values but you are getting rid off them with the way you are working through your answer by moving the exponent of x down as then you have \ln(x) and x cannot be negative for this expression, while x can be negative in \ln(x^2) as negative numbers squared become positive, so you'd be taking the log of a positive number which is valid hence negative x values are valid.
    • Thread Starter
    Online

    2
    ReputationRep:
    How can I start to find \sum^n_{r=1} r10^{-r} in terms of n?

    That is, just drop me a hint and not the full solution.
    Offline

    3
    ReputationRep:
    (Original post by Palette)
    How can I start to find \sum^n_{r=1} r10^{-r} in terms of n?

    That is, just drop me a hint and not the full solution.
    Differentiate the geometric series.
    • Thread Starter
    Online

    2
    ReputationRep:
    (Original post by Zacken)
    Differentiate the geometric series.
    The derivative of 10^{-x} gives -(\ln 10)10^{-x} but I don't know how it will help.

    Would this differentiation approach work for any \sum^n_{r=1} f(r)g(r) where g(r) is a geometric series?
    Offline

    3
    ReputationRep:
    (Original post by Palette)
    The derivative of 10^{-x} gives -(\ln 10)10^{-x} but I don't know how it will help.

    Would this differentiation approach work for any \sum^n_{r=1} f(r)g(r) where g(r) is a geometric series?
    Not what I meant.

    \displaystyle f(k) =  \sum_{r=0}^n k^r \Rightarrow \frac{df}{dk} =  \sum_{r=1}^n rk^{r-1}= k^{-1}\sum_{r=1}^n rk^r or along these lines w/e

    and no to your last question
    • Thread Starter
    Online

    2
    ReputationRep:
    (Original post by Zacken)
    Not what I meant.

    \displaystyle f(k) =  \sum_{r=0}^n k^r \Rightarrow \frac{df}{dk} =  \sum_{r=1}^n rk^{r-1}= k^{-1}\sum_{r=1}^n rk^r or along these lines w/e

    and no to your last question
    I don't want to sound stupid, but why can we differentiate f(k) when r can only take positive integer values between [0 and n?
    Offline

    3
    ReputationRep:
    (Original post by Palette)
    I don't want to sound stupid, but why can we differentiate f(k) when r can only take positive integer values between [0 and n?
    Simplified, your question is isomorphic to: "Why can we differentiate x^2 when 2 is a positive integer?"

    I explicitly pointed out that we're differentiating with respect to k (a variable that can vary over the reals) by writing df/dk and not f'(k). We're not differentiating with respect to r, it's just a constant.
    • Thread Starter
    Online

    2
    ReputationRep:
    (Original post by Zacken)
    Simplified, your question is isomorphic to: "Why can we differentiate x^2 when 2 is a positive integer?"

    I explicitly pointed out that we're differentiating with respect to k (a variable that can vary over the reals) by writing df/dk and not f'(k). We're not differentiating with respect to r, it's just a constant.
    I get it now- thanks for your help!
    • Thread Starter
    Online

    2
    ReputationRep:
    (Original post by Zacken)
    Simplified, your question is isomorphic to: "Why can we differentiate x^2 when 2 is a positive integer?"

    I explicitly pointed out that we're differentiating with respect to k (a variable that can vary over the reals) by writing df/dk and not f'(k). We're not differentiating with respect to r, it's just a constant.
    Having looked at my question again, I realized how silly it was.
    Offline

    3
    ReputationRep:
    (Original post by Palette)
    Having looked at my question again, I realized how silly it was.
    The original one or the "r constant" one?

    By the way, even if r could take on positive integer values, you could still take on r as a real variable, differentiate with respect to it and then send it back to the integer space in nice enough conditions and/or sufficient handwaving.
    Online

    2
    ReputationRep:
    (Original post by Palette)
    Having looked at my question again, I realized how silly it was.
    Same - i only realise when its pointed out to me. Im just glad im anonymous on this site aha

    - thanks to the people who helped me on that ln question. I think i understand now
    Offline

    3
    ReputationRep:
    (Original post by k.russell)
    now I am a future biologist, so maybe not a leading authority on engineering - but I do think I've worked it out.
    It makes sense the block will reach terminal velocity, right, when the forces acting forward = those acting backward.
    So you set up the diagram, mg sin(a) down the slope and uN + kV up the slope, as we are adding in the air resistance term (kV). N = mg cos (a) , so you end up with this;
    mg sin(a) = u mg cos(a) + kV
    and you rearrange it get V out, because that V is gonna be the terminal velocity. At lesser values of V, the forces down slope would be greater and the object would accelerate. Do you agree with that?? Hope you can understand it lol (I had to use a for alpha and u for mu)
    Yeah that makes sense thanks a lot, forgot to include a=0 when it's at constant velocity ie terminal velocity
    Offline

    2
    ReputationRep:
    (Original post by metrize)
    Yeah that makes sense thanks a lot, forgot to include a=0 when it's at constant velocity ie terminal velocity

    lol np, I am glad (and surprised) I could help out I am guessing you are studying some M units, how many have you done so far?
    Offline

    2
    ReputationRep:
    Sorry to intrude, but just out of interest, I'd like to ask those who do both physics and further maths which is the more difficult? (I do further maths, not physics)


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by RedSquirrels)
    Sorry to intrude, but just out of interest, I'd like to ask those who do both physics and further maths which is the more difficult? (I do further maths, not physics)


    Posted from TSR Mobile
    Further maths can be very easy if you're interested in maths.
    Online

    3
    ReputationRep:
    (Original post by RedSquirrels)
    Sorry to intrude, but just out of interest, I'd like to ask those who do both physics and further maths which is the more difficult? (I do further maths, not physics)


    Posted from TSR Mobile
    I think Physics. You can choose fairly 'easy' (as in work hard, practice a lot, gain understanding and ace the exam) modules and it can be almost... straightforward.

    Physics doesn't test your mathematical abilities as much but there's a lot of theory, used to have oracticals as well which was a nightnare, and like a lot of science subjects the exams/markschemes are quite annoying about things.
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: November 26, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Today on TSR
Poll
Would you rather have...?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read here first

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.