Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    17
    ReputationRep:
    Log question - express in terms of logax , logay, logaz

    Ive attached the ques h.) & my working but i dont understand where the 5/2 comes from.

    Answer is 5/2 + 1/2logax - 3logay + 5/2logaz

    Posted from TSR Mobile
    Attached Images
      
    Offline

    22
    ReputationRep:
    di
    (Original post by kiiten)
    Log question - express in terms of logax , logay, logaz

    Ive attached the ques h.) & my working but i dont understand where the 5/2 comes from.

    Answer is 5/2 + 1/2logax - 3logay + 5/2logaz

    Posted from TSR Mobile
    \log_2 4 = \log_2 2^2 = 2\log_2 2 = 2 \times 1 = 2

    \displaystyle \log_a \frac{\sqrt{axz^5}}{y^3} = \log_a \sqrt{axz^5} - \log_a y^3

    Now \log_a y^3 = 3\log_a y

    and \log_a \sqrt{axz^5} = \frac{1}{2} \log_a (axz^5) = \frac{1}{2}\left(\log_a a + \log_a x + \log_a z^5\right)

    Can you take it from here? What's \log_a a?
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    di

    \log_2 4 = \log_2 2^2 = 2\log_2 2 = 2 \times 1 = 2

    \displaystyle \log_a \frac{\sqrt{axz^5}}{y^3} = \log_a \sqrt{axz^5} - \log_a y^3

    Now \log_a y^3 = 3\log_a y

    and \log_a \sqrt{axz^5} = \frac{1}{2} \log_a (axz^5) = \frac{1}{2}\left(\log_a a + \log_a x + \log_a z^5\right)

    Can you take it from here? What's \log_a a?
    How did you get to -logay^3 on the second line of working ?
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    How did you get to -logay^3 ?
    \log \frac{a}{b} = \log a - \log b except in this case, we have a =\sqrt{axz^5} and b = y^3
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    \log \frac{a}{b} = \log a - \log b except in this case, we have a =\sqrt{axz^5} and b = y^3
    Umm i think ive done it wrong but i got
    2 + 1/2loga + 1/2logx + 5/2logz - 3logay
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    Umm i think ive done it wrong but i got
    2 + 1/2loga + 1/2logx + 5/2logz - 3logay
    Yeah, that's correct, just remember that \frac{1}{2} \log_a a  = \frac{1}{2} \times 1 = \frac{1}{2} so you can simplify your first two terms as 2 + \frac{1}{2} = \cdots
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    Yeah, that's correct, just remember that \frac{1}{2} \log_a a  = \frac{1}{2} \times 1 = \frac{1}{2} so you can simplify your first two terms as 2 + \frac{1}{2} = \cdots
    So 5/2 + 1/2logax + 5/2logaz - 3logay
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    So 5/2 + 1/2logax + 5/2logaz - 3logay
    Seems fine, yups.
    • Thread Starter
    Offline

    17
    ReputationRep:
    Thank you

    If you have log (1+5^-x) would it simplify to -xlog6 ?
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    Thank you

    If you have log (1+5^-x) would it simplify to -xlog6 ?
    No, it wouldn't; you can't simplify that anymore. Remember that \log(a+b) \neq \log a+ \log b. :-)
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    No, it wouldn't; you can't simplify that anymore. Remember that \log(a+b) \neq \log a+ \log b. :-)
    So how would you solve
    Log (1×5^-x) = log3
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    So how would you solve
    Log (1×5^-x) = log3
    "cancel" the logs: 1 + 5^{-x} = 3 \Rightarrow 5^{-x} = 2 \Rightarrow 5^x = \frac{1}{2} - now you know how to solve this using logarithms.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    "cancel" the logs: 1 + 5^{-x} = 3 \Rightarrow 5^{-x} = 2 \Rightarrow 5^x = \frac{1}{2} - now you know how to solve this using logarithms.
    Oh i see - could you do that for any log e.g. log5^-x = log3 instead of simplifying
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    Oh i see - could you do that for any log e.g. log5^-x = log3 instead of simplifying
    If you have \log a = \log b - then you know that a=b.

    If you have \log a + \log b = \log c IT IS NOT TRUE THAT a+b = c BUT IT IS TRUE that \log a + \log b = \log ab = \log c \Rightarrow ab=c

    Basically when you have log equations you should try and get it in the form \log (\text{something}) = \log (\text{something else}) then you can say something = something else. (the bases of the logs have to be the same, if they're not, you can always use change of base rule) .
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Zacken)
    If you have \log a = \log b - then you know that a=b.

    If you have \log a + \log b = \log c IT IS NOT TRUE THAT a+b = c BUT IT IS TRUE that \log a + \log b = \log ab = \log c \Rightarrow ab=c

    Basically when you have log equations you should try and get it in the form \log (\text{something}) = \log (\text{something else}) then you can say something = something else. (the bases of the logs have to be the same, if they're not, you can always use change of base rule) .
    Thanks that helps to explain things
    Offline

    22
    ReputationRep:
    (Original post by kiiten)
    Thanks that helps to explain things
    No problem.
    • Thread Starter
    Offline

    17
    ReputationRep:
    What about this question - im not sure where you would start. Ive attached some workingAttachment 513103513107

    Name:  1458040767458607763651.jpg
Views: 87
Size:  498.4 KB
    Attached Images
     
    Offline

    21
    ReputationRep:
    Name:  image.png
Views: 64
Size:  51.3 KB

    Use Log rules.
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by Mystery.)
    Name:  image.png
Views: 64
Size:  51.3 KB

    Use Log rules.
    Have i done it right so far. So, 5/ (log3x) = 6 / (log3)
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by kiiten)
    What about this question - im not sure where you would start. Ive attached some workingAttachment 513103513107

    Name:  1458040767458607763651.jpg
Views: 87
Size:  498.4 KB
    You have

    \log_3 x + log_3 y = 5

    \log_3 x \times log_3 y = 6


    Let a=\log_3 x and b=\log_3 y then you have

    a+b=5

    ab = 6

    So you can ignore logs and solve these equations for a and b. Then substitute back in for a and b.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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.