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Logarithmic equation

  1. Jul 3, 2009 #1
    log9 (p) = log12 (q) = log16 (p+q)
    What is q/p?

    9^x + 12^x = 16^x
    calculate x
    (4/3)^x = q/p

    I guess that would work if I'd know how to solve that equation (without using a calculator), unfortunately I don't...

    3^(2x)+4^x*3^x = 4^2x
     
  2. jcsd
  3. Jul 3, 2009 #2
    Divide by 4^x 3^x ... Can you solve a + 1 = 1/a ?
     
  4. Jul 3, 2009 #3
    Thanks! :-)

    (4/3)^-x + 1 = (4/3)^x

    substitute (4/3)^x by y

    y^-1 + 1 = y^1

    multiply by y

    1 + y - y^2 = 0

    (1/2) * (1 + sq(5) ) = (4/3)^x = q/p
    (1 - sq(5) < 0 , not a possible solution)
     
  5. Jul 4, 2009 #4

    What the heck is a+1=1/a? I can't get an answer. I get to (a squared + a)=1, then I'm stuck.

    Steve
     
  6. Jul 4, 2009 #5

    Hootenanny

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    Gold Member

    Try taking a factor of a out of the LHS ... :wink:
     
  7. Jul 4, 2009 #6
    quadratic equation
     
  8. Jul 4, 2009 #7
    I do not understand what is the original task. Could you possibly rewrite it?
     
  9. Jul 4, 2009 #8
    Sure :)
    (p en q > 0) :
    log9(p) = log12(q) = log16(p + q) :
    What is q/p?
     
  10. Jul 5, 2009 #9
    If log9*(p)=log12*(q) then 9p=12q and q/p = 9/12
     
  11. Jul 5, 2009 #10
    log9(p) = log12(q) = log16(p + q)
    is
    9^x + 12^x = 16^x
    (9, 12 and 16 are the bases eh)

    If you solve that equation you get:
    q/p = (1/2) * (1 + sq(5) )

    I'm wondering, if 9 and 12 were coefficients, would this be the solution?

    9 log(p) = 12 log(q)
    log(q) = 1 so q = 10
    log(p) = 12/9 so p = 10^(12/9)

    q/p = 10^(1-12/9) = 10 ^ (-1/3)
     
  12. Jul 5, 2009 #11

    HallsofIvy

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    What reason would you have to think that log(q)= 1?
    If 9 log(p)= 12 log(q) then log(p)/log(q)= 12/9= 4/3. You could then say that log(p)/log(q)= logq(p)= 4/3 so that p= q4/3 and p/q = q4/3. There are many possible values for p/q.

    For example, if q= 8, p= 16, then 9log(p)= 9log(16)= 9 log(16)= 9log(2^4)= 9(4)log(2)= 36 log(2)= 12(3)log(2)= 12log(2^3)= 12log(8)= 12log(q). So 9 log(p)= 12 log(q) and p/q= 16/8= 2.

    But if q= 27, p= 81, then 9log(p)= 9log(81)= 9log(3^4)= 36log(3)= 12(3)log(3)= 12log(33= 12 log(27)= 12 log(q). Again 9 log(p)= 12 log(q) but now p/q= 81/27= 3.

     
  13. Jul 5, 2009 #12
    If [itex]log_9(p)=log_{12}(q)=log_{16}(p+q)[/itex] then

    [tex]9^x + 12^x = 16^x[/tex]

    [tex]y=(\frac{4}{3})^x[/tex]

    [tex]y_{1,2}=\frac{1 \pm \sqrt{5}}{2}[/tex]

    If:

    [tex](\frac{4}{3})^x=\frac{1 \pm \sqrt{5}}{2}[/tex]

    Then what is x?

    [tex]x=log_\frac{4}{3}{\frac{1 + \sqrt{5}}{2}}[/tex]

    You would tell me why [tex]y_2=\frac{1 - \sqrt{5}}{2}[/tex] is not the solution for [tex](\frac{4}{3})^x[/tex]

    Regards.
     
    Last edited: Jul 5, 2009
  14. Jul 5, 2009 #13
    Because it's a negative number, but I already figured out myself in post #3, thanks anyway ;-)
     
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