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

    2
    ReputationRep:
    Please explain how to sketch the graph of y=ln(2x-3). Thanks.
    Offline

    14
    ReputationRep:
    (Original post by Sister Clo)
    Please explain how to sketch the graph of y=ln(2x-3). Thanks.
    Start by looking at the equations

    y=ln(x)
    y=ln(2x-3)

    What transformation/s turn ln(x) into ln(2x-3)

    Then apply these to the graph

    It's also a good idea to sub in y=0 or x=0 and solve for x or y respectively to find some intercepts to guide you
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by Sister Clo)
    Please explain how to sketch the graph of y=ln(2x-3). Thanks.
    Consider what transformations take place when you go from \ln(x) (which I hope you know how to draw) to \ln(2x-3). Then consider how it would look like after these transformations, keeping in mind the critical points.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    Consider what transformations take place when you go from \ln(x) (which I hope you know how to draw) to \ln(2x-3). Then consider how it would look like after these transformations, keeping in mind the critical points.
    Thanks. I transform by 2x (stretch in the x direction, so half the x-values), then translate 3 to the right. Doesnt seem to be the same when using a grapher.
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by Sister Clo)
    Thanks. I transform by 2x (stretch in the y direction), then translate 3 to the right. Doesnt seem to be the same when using a grapher.
    Why the y-direction? You're applying these onto y=\ln(x), not y=x. If it were the latter than it would be true, however alternatively it is also a stretch by s.f. 1/2 in x-direction.

    Before that happens though, I recommend you translate it by 3 to the right because \ln(x) \mapsto \ln(x-3) is a translation to the right by 3 units, and then \ln(x-3) \mapsto \ln(2x-3) is a stretch by s.f. 1/2 along the x-axis.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by AHappyStudent)
    Start by looking at the equations

    y=ln(x)
    y=ln(2x-3)

    What transformation/s turn ln(x) into ln(2x-3)

    Then apply these to the graph

    It's also a good idea to sub in y=0 or x=0 and solve for x or y respectively to find some intercepts to guide you
    Thanks - I stretch in the x-direction by scale factor half then translate 3 to the right. Seems to be different using graph plotter
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    Why the y-direction? You're applying these onto y=\ln(x), not y=x. If it were the latter than it would be true, however alternatively it is also a stretch by s.f. 1/2 in x-direction.

    Before that happens though, I recommend you translate it by 3 to the right because \ln(x) \mapsto \ln(x-3) is a translation to the right by 3 units, and then \ln(x-3) \mapsto \ln(2x-3) is a stretch by s.f. 1/2 along the x-axis.
    Thanks yet again - I think the issue I have problem getting my head around is why translate first, when I would do the 2x first for all other graphs without a problem.
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by Sister Clo)
    Thanks yet again - I think the issue I have problem getting my head around is why translate first, when I would do the 2x first for all other graphs without a problem.
    You can do stretch first, just depends on the preference. You apply transformations to x, so if you do stretch by s.f. 1/2 first you end up with \ln(2x) and now you need to translate by an appropriate a such that \ln(2(x-a)) is equal to \ln(2x-3) - clearly a\neq 3 here like it could be in the first scenario.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by RDKGames)
    You can do stretch first, just depends on the preference. You apply transformations to x, so if you do stretch by s.f. 1/2 first you end up with \ln(2x) and now you need to translate by an appropriate a such that \ln(2(x-a)) is equal to \ln(2x-3) - clearly a\neq 3 here like it could be in the first scenario.


    Thank you. Much appreciated.
 
 
 
  • 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
    What's your favourite Christmas sweets?
    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.