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Log question (HL maths)

  • 08-06-2011 9:13pm
    #1
    Registered Users Posts: 204 ✭✭


    Okay so I wanted to do question 5 as a back up on Friday. What's everyone expecting to come up on it?

    Also: How the heck do you do change of base law?

    Here is my problem:

    Show that log[base3](2-3x) = 2log[base9](2-3x) :confused:


Comments

  • Closed Accounts Posts: 494 ✭✭PJelly


    Bring the two over the second log so its squared.
    Change the base on the right side so you get Log(3)[2-3x]^2 (all over) Log(3)[9]
    which equals that previous log over two.

    Multiply across by that two and on the left you get 2log[base3](2-3x) which is equal to Log(base3)[2-3x]^2
    Which equals the right hand side.


  • Registered Users Posts: 204 ✭✭polka dot


    You are a life saviour! Thank you so much. :D I've spent the last half an hour trying to figure out how to do "log(3)9" on my calculator before I managed to get it by sheer dumb luck...

    Question 5 is obviously going to go well.


  • Registered Users, Registered Users 2 Posts: 372 ✭✭Patriciamc93


    polka dot wrote: »
    Okay so I wanted to do question 5 as a back up on Friday. What's everyone expecting to come up on it?

    Also: How the heck do you do change of base law?

    Here is my problem:

    Show that log[base3](2-3x) = 2log[base9](2-3x) :confused:

    Don't forget page 21 of log tables. They really are a major help!


  • Closed Accounts Posts: 494 ✭✭PJelly


    Oh and this is how I remember change of base law
    Log of the number (not the base) on top
    Log of the base on the bottom of the fraction

    Now set them both to any base

    Like... (assume brackets means base) Log(2)3
    Equals Log(anything)3 All over Log(anything, but must be same as above)2

    And, lets assume we change to base 5, and / denotes a fraction.
    Log(8)3 = Log(5)3/Log(5)8

    Geddit?


  • Closed Accounts Posts: 494 ✭✭PJelly


    polka dot wrote: »
    You are a life saviour! Thank you so much. :D I've spent the last half an hour trying to figure out how to do "log(3)9" on my calculator before I managed to get it by sheer dumb luck...

    Question 5 is obviously going to go well.
    Oh I just did it in my head. Just think, it's saying "What power do I put three to, to get nine?"
    Answer is two. 3^2 is nine ^_^


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  • Registered Users Posts: 204 ✭✭polka dot


    PJelly wrote: »
    Oh I just did it in my head. Just think, it's saying "What power do I put three to, to get nine?"
    Answer is two. 3^2 is nine ^_^

    Everything makes so much more sense when you think of it that way. :rolleyes:

    Is the question meant to be more binomial expansions this year or more logs/everything else?

    Thank you everyone. :)


  • Closed Accounts Posts: 494 ✭✭PJelly


    polka dot wrote: »
    Everything makes so much more sense when you think of it that way. :rolleyes:

    Is the question meant to be more binomial expansions this year or more logs/everything else?

    Thank you everyone. :)
    I honestly have no idea.
    Not including proofs, I think its basically impossible to predict a maths exam


  • Registered Users Posts: 57 ✭✭conlufc


    PJelly wrote: »
    Oh and this is how I remember change of base law
    Log of the number (not the base) on top
    Log of the base on the bottom of the fraction

    Now set them both to any base

    Like... (assume brackets means base) Log(2)3
    Equals Log(anything)3 All over Log(anything, but must be same as above)2

    And, lets assume we change to base 5, and / denotes a fraction.
    Log(8)3 = Log(5)3/Log(5)8

    Geddit?
    I understand it......just
    Q5 isnt da worst...logs and proof by induction would b nice


  • Registered Users Posts: 204 ✭✭polka dot


    PJelly wrote: »
    I honestly have no idea.
    Not including proofs, I think its basically impossible to predict a maths exam

    Are there proofs for this question?


  • Closed Accounts Posts: 494 ✭✭PJelly


    Yeah, a fraction with logs that ends in 1/log(r!)30


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