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Homework Help: Simultaneous equations help

  1. Feb 14, 2006 #1
    I need some help with this please!

    1. given p=logq 16, express in terms of p:
    a) logq 2
    b) logq (8q)

    this is what i got:
    a) p= logq16 = logq22
    = 4logqqq
    =1/4 p

    b) p=logq16 =logq8 +logq q
    =logq23 + 1
    =3logq2 +1

    Also can someone please help get me started on these similtaneous equations?

    8y =42x+3
    log2 y = log2x +4

    I'm not sure exactly how to equate them
  2. jcsd
  3. Feb 14, 2006 #2
    For the first one, the answer ([itex]\log_q2=p/4[/itex]) is OK, but I have trouble understanding what you wrote before that.

    For the second one, you can't assume that 16=8q. Otherwise, you're on the right track:


    You already know what [itex]\log_q2[/itex] is in terms of p, so you're done.


    For the third one, try to isolate y in terms of x from the second equation. Remember that [itex]\log_216=4[/itex]. Then substitute that expression into the first equation and rewrite the eq. as a logarithm in base 8. Solve for x and then y.

    EDIT2: Tnx, arildno, no wonder I didn't understand what he wrote, since it's wrong.

    - Kamataat
    Last edited: Feb 15, 2006
  4. Feb 15, 2006 #3


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    I'm not quote sure what you are doing here. But since, 8 = 23, and 4 = 22. So taking the logarithm base 2 of both sides, gives:
    8y = 42x + 3
    <=> log2(8y) = log2(42x + 3)
    <=> log2(23y) = log2(22(2x + 3))
    You can go from here, right? :)
  5. Feb 15, 2006 #4


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    Dearly Missed

    a) is totally wrong.
    You are in effect saying p=1/4p, which is incorrect.
  6. Feb 16, 2006 #5
    ok i'm sorry for not setting it out clearly enough. What i meant was:
    log 9 2 = 1/4 p
    which according to my text book is correct

    don't worry i've managed to work out b) as well
    if log 9 2 = 1/4 p
    then logq(8q) = logq (23q)
    so logq(8q) = 3/4p +1
    Last edited: Feb 16, 2006
  7. Feb 16, 2006 #6


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    Have you worked out the equation one yet?
    Is it clear, or do you need more hints? :)
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