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Help with a log

  1. Jul 26, 2006 #1
    Hello,

    I'm having trouble recovering all the solutions for this equation; here is the work I've managed to produce thus far.

    logx=\sqrt{logx}

    (logx)^2=logx

    10^logx^2=x <--I'm not sure of this move?

    x^2=x

    x^2-x=0

    x(x-1)=0

    x= 1, 0

    I know 0 cannot work, and 10 looks likes a possible solution, but I cannot produce it--anyways, any help would be appreciated.

    thanks.
     
    Last edited: Jul 26, 2006
  2. jcsd
  3. Jul 26, 2006 #2

    Tide

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    Why don't you try setting y = log x in the original equation then solve for y? That will avoid some confusion.
     
  4. Jul 26, 2006 #3
    y=logx

    x=10^(logx)^1/2

    This is where I'm stuck, because I feel it would be illegal to assume

    10^logx^1/2...shouldn't it be (logx)^1/2. I don't know how to simplify after this point.
     
  5. Jul 26, 2006 #4
    Apply Tide's hint to this equation: [tex](logx)^2=logx[/tex]

    Once you have solved for y, replace y with logx, and then solve for x.
     
  6. Jul 26, 2006 #5

    VietDao29

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    Uhmm, no:
    (log x)2 = log(x)
    By letting y = log(x), we have:
    y2 = y
    <=> y2 - y = 0
    And this is our familiar quadratic equation in y. Can you solve for y from that equation? Then from there, just solve for x.
    Can you go from here? :)
     
  7. Jul 26, 2006 #6
    Ahh, thanks! I can't believe I couldn't think of this, haha.

    (logx)^2-logx=0

    logx(logx-1)=0

    logx=0 , log(x-1)=0

    x=1, 10
    A quick two questions: is there any other way to solve this, other than the y^2-y=0, and still maintain solutions 1, 10? For (logx)^n n>1, how would you treat this in an equation--merely a^n=(logx)^n never expanding?

    Thanks,
    Chris
     
  8. Jul 26, 2006 #7

    VietDao29

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    Yes, this is correct. Congratulations. :smile:
    Uhm... no.
    By the way, this is the common way, and is fast enough. So, I don't think you'll need some other ways though. :)
    Yes, if it helps you visualize better. Say, it would be much better if you substitute y = log(x), and all of sudden, the whole equation becomes a quadratic equation, or a cubic equation,... in y. Right?
    Normally, to solve an equation, we simplify the equation first to make it look less complicated. Then we isolate x to one side of the equation. Then we solve it from there.
    Of course, the equation:
    [tex]x = 10 ^ {\sqrt{\log x}}[/tex] just looks horible, right?
    You're welcome. :)
     
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