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Changing the bases of logs question!

  1. Mar 16, 2013 #1
    1. The problem statement, all variables and given/known data

    solve the equation log4 x = 1 + log 2 2x , x>0




    My attempt

    I changed it base 2

    so I now have( when i write log it's to the base 2)

    log2x/( log2 4) = log 2 + log 2x

    and I'm stuck here I keep trying to form an equation, but I can't.
     
  2. jcsd
  3. Mar 16, 2013 #2

    trollcast

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    Use the log identities to take get one log on the right hand side.

    Then since log_2(4) equals two you can multiply by 2 to get rid of the fraction on the left hand side.

    Then log identities again to get it so you just have 2 logs. And then it should be simple from there on.
     
  4. Mar 16, 2013 #3
    Yeah I think I see it now :S

    2^x = 2^-4

    = 1/16
     
  5. Mar 16, 2013 #4

    trollcast

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    Any chance you could post more of your solution as I can't see how you're getting to that step (but log questions can have a couple of possible solutions), although your answer is right.
     
  6. Mar 16, 2013 #5
    Umm logx/log4 = log 2 + log 2 + log x

    logx = 4 + 2logx

    2logx-logx +4 =0

    logx = log2^-2
    ..

    Seems my working is wrong I'm stuck again
     
  7. Mar 16, 2013 #6

    trollcast

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    Go back to the equation and see you can get it from there:

    $$\frac{log_2 (x)}{log_2(4)} = log_2(2)+log_2(2x)$$

    Now use the log identity, [itex]log_b(xy)=log_b(x)+log_b(y)[/itex] to simplify the right hand side.

    After you do that can you see how to get rid of the fraction on the left hand side?
     
  8. Mar 16, 2013 #7
    Uhh I think i really got it this time what I did now was

    (log x)/ (log4) = log 2 + log 2 + log x

    (logx)/(log4) = 2 + log x

    (logx)/2 = 2+ log x

    logx = 4 + 2logx

    2logx-logx +4 = 0

    logx = log 2^-4

    logx = log(1/16)

    x= 1/16
     
  9. Mar 16, 2013 #8

    trollcast

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    Yeah thats right.
     
  10. Mar 16, 2013 #9
    Is that the same method you were telling me to try?
     
  11. Mar 16, 2013 #10

    trollcast

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    Yeah its pretty much the same (as I said earlier theres usually a couple of different routes to the same answer with a log question), starting from the line in your OP

    $$\frac{log_2(x)}{2}=log_2(4x)\\
    \\log_2(x)=log_2(16x^2)
    \\x=16x^2
    \\x(16x -1) = 0$$
    [itex]∴ x = \frac{1}{16}[/itex] as solution x = 0 is not possible.
     
  12. Mar 16, 2013 #11
    oh lol your way is much simpler =.= . I'm so blind D:
     
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