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Homework Help: Logarithm equation

  1. Apr 9, 2005 #1
    hi

    could anyone tell me where I went wrong ?

    simultaneously solve

    2logbase2 y = logbase4 3 + logbase2 x

    3^y = 9^x

    But for the top I get y = 3 root x

    and bottom I get y=3x

    so what's gone wrong ?


    thanks

    roger
     
  2. jcsd
  3. Apr 9, 2005 #2

    dextercioby

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    Use [itex] \ln [/itex]

    The second becomes

    [tex] y\ln 3=2x\ln 3\Rightarrow y=2x [/tex] (1)

    while the first

    [tex] 2\ln y=\frac{\ln 3}{2}+\ln x [/tex] (2)

    Solve the simple system (1) + (2)

    Daniel.
     
  4. Apr 9, 2005 #3
    do you mean log base e when you state ln ?

    But my working out is as follows and please tell me what went wrong :

    logbase2 y ^2 = logbase2 root3 + logbase2 x

    = log base 2 xroot3 = log base2 y ^2
    then remove logs and square both sides to get y = 3 root x
     
  5. Apr 9, 2005 #4

    HallsofIvy

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    Sorry, but I can't find anything that's gone right.

    9= 32 so 9x= 32x.

    The second equation is 3y= 32x which gives y= 2x, not y= 3x.

    The first equation is 2 log2(y)= log4(3)+ log2(x)
    log2(y2)- log2(x)= log4(3)
    log2(y2/x)= log4(3)

    If z= log4(3) then 3= 4z= (22)z= 22z. Now taking log2 of both sides, gives log2(3)= 2z so log4(3)= (1/2)log2(3)= log2(31/2).

    Putting those together, log2(y2/x)= log2(31/2) so that y2/x= 31/2 or y= 31/4x1/2, not y= 3 x1/2.

    That is, we have 2x= 31/4x1/2.

    One obvious answer to that is x= 0.

    Squaring both sides, 3x2= 31/2x and if x is not 0,
    4x= 31/2 so x= 31/2/4 is another solution.
     
    Last edited by a moderator: Apr 9, 2005
  6. Apr 9, 2005 #5

    dextercioby

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    Halls,there's no such thing as logarithm from 0...That "x" equal 0 doesn't satisfy the first equation.

    Daniel.
     
  7. Apr 9, 2005 #6
    why ?

    and dextercioby, x cant be negative but cant it tend to zero to give the power of negative infinity ?
     
    Last edited: Apr 9, 2005
  8. Apr 9, 2005 #7

    dextercioby

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    No,no.That is x=0...It's an exact solution.It's not acceptable,as 0 is not in the domain of logarithm (in any base).

    Daniel.
     
  9. Apr 10, 2005 #8

    HallsofIvy

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    dexercioby was right, I was wrong- I forgot to check my answer in the original equation. The problem did not ask about limits, it asked about values for specific x. Since log2(0) is not defined, x= 0 cannot satisfy the equation.

    [tex]x= \frac{\sqrt{3}}{4}, y= \frac{\sqrt{3}}{2}[/tex] is the only solution.
     
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