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Math Help - logs again

  1. #1
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    logs again

    I really am trying to understand this. According to your last advice, letting y=ln(x), see if I am getting any closer.

    lnx = 2(ln3-ln5)
    y = 2(3y-5y)
    y = -2y * 2

    so (hopefully)
    ln(x) = -2e^2

    or is it ln(x) = 2e^-2
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by startingover View Post
    I really am trying to understand this. According to your last advice, letting y=ln(x), see if I am getting any closer.

    lnx = 2(ln3-ln5)
    y = 2(3y-5y)
    y = -2y * 2

    so (hopefully)
    ln(x) = -2e^2

    or is it ln(x) = 2e^-2
    i'm sure no one here gave you such advice for a problem of this nature. use the laws of logarithms to write the right side as a single log, then just equate what's being logged.

    lnx = 2(ln3 - ln5)
    => lnx = 2(ln(3/5)) ...........since logx - logy = log(x/y)
    => lnx = ln(3/5)^2 ...........since log(x^n) = nlogx
    => lnx = ln[(3/5)^2]
    => x = (3/5)^2
    => x = 9/25

    besides, even if you wanted to make lnx = y to make things look easier, you would not change ln3 and ln5 into y since they are not lnx, so going your way, you would end up with:

    y = 2(ln3 - ln5) ........doing the same operations as above, you'd end up with
    y = ln[(3/5)^2]
    but y = lnx
    => lnx = ln[(3/5)^2]
    => x = 9/25
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  3. #3
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    Now I am confused

    Quote:
    Originally Posted by startingover
    5= 3lnx - 1/2lnx

    I assume you have to get both sides of the equation with the same term first. If I multiply by lnx, wouldn't it cancel out on the right side? The same for if I divide?

    Look at it this way:
    Let y = ln(x)
    Then:
    5 = 3y - (1/2)y

    5 = (5/2)y

    y = 5*2/5 = 2

    ln(x) = 2

    x = e^2

    -Dan
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by startingover View Post
    Quote:
    Originally Posted by startingover
    5= 3lnx - 1/2lnx

    I assume you have to get both sides of the equation with the same term first. If I multiply by lnx, wouldn't it cancel out on the right side? The same for if I divide?

    Look at it this way:
    Let y = ln(x)
    Then:
    5 = 3y - (1/2)y

    5 = (5/2)y

    y = 5*2/5 = 2

    ln(x) = 2

    x = e^2

    -Dan
    ok, so you notice you had 2 lnx's here, in the problem you are doing now, you only have one. if you let y = lnx, you would only change the left side. apparently Dan thought you would have a problem moving lnx around so he made it easier on you by replacing it with a variable, but in general this is not necessary, at least for a problem like this:

    5 = 3lnx - 1/2lnx
    => 5 = ln(x^3) - ln(x^(1/2))
    => 5 = ln[(x^3)/(x^(1/2))]
    => 5 = ln[x^(5/2)]
    => e^5 = x^(5/2)
    => (e^5)^(2/5) = x
    => e^2 = x

    or you can do it Soroban's way, which i like better, it's not as complicated

    you see here that doing it without changing lnx to y makes the problem a bit more challenging, but it still was not necessary. in any case, this is not a problem like the one you have here, on the current problem, there is only one variable ln(x), in the above problem, all variables are lnx, so making a change might make sense if you want a little simplier algebra. so i am still correct about no one telling you that advice for a problem of this nature

    usually, we use Dan's technique with slightly harder problems such as quadratics where the variable is a log or another complicated function. example:

    (lnx)^2 - 5lnx - 6 = 0 ...............this looks hard, but let y = lnx and we get:
    y^2 - 5y - 6 = 0 ......................this looks easy, it's just your old friend the quadratic, but you do the same manipulations
    Last edited by Jhevon; April 22nd 2007 at 08:37 AM.
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