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Homework Help: Help with logarithms

  1. Jan 3, 2010 #1
    I have [tex]2lnx = xln2[/tex]
    where [tex]x\ne2[/tex]

    if you start by dividing both sides by ln2
    is the following legal?

    [tex]\frac{2lnx}{ln2} \rightarrow x = 2ln(x-2)[/tex]

    [tex]e^{2ln(x-2)} = (x-2)^2[/tex]

    [tex]x = (x-2)^2 \implies x = 4[/tex]
     
  2. jcsd
  3. Jan 3, 2010 #2

    Mentallic

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    I'm confused at what you've done.

    Did you manipulate [tex]\frac{2lnx}{ln2}[/tex] and turn it into [tex]2ln(x-2)[/tex]? Because this is generally not correct.

    From what I can gather, you've just fluked your way into breaking a rule but still finding a solution for x. That equation isn't simple to solve.
     
  4. Jan 3, 2010 #3
    Hm,

    [tex]\frac{ln~x}{ln~y} \implies ln(x-y)[/tex]

    ?
     
  5. Jan 3, 2010 #4


    [tex]2lnx = xln2\Rightarrow \frac{d[2lnx]}{dx}=ln2\frac{dx}{dx}\Rightarrow \frac{2}{x}=ln2\Rightarrow x=\frac{2}{ln2}[/tex]
     
    Last edited: Jan 3, 2010
  6. Jan 3, 2010 #5
    Hm, no!
     
  7. Jan 3, 2010 #6
    Then tell me what did I do wrong. You follow step by step what I did?
     
    Last edited: Jan 3, 2010
  8. Jan 3, 2010 #7
    No, you have it completely backwards... :smile:
    [tex]\ln\left(\frac{x}{y}\right) = \ln x - \ln y[/tex]

    but this isn't going to help actually solve the equation.
     
  9. Jan 3, 2010 #8

    Redbelly98

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    Hint: review the rules for manipulating logs. One of the says that
    A ln(B) = ____?​
     
  10. Jan 3, 2010 #9
    I got fool too!!!! And I destroy all the evidence already!!!:rofl::tongue:


    I did it differently and I still don't see why he claimed my answer in #4 was wrong!!!
     
  11. Jan 3, 2010 #10

    Mentallic

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    He claimed it was wrong because it is wrong...

    You took the derivative and solved for that. Yeah you found the turning point, but not a root of the original equation.
     
  12. Jan 3, 2010 #11

    ideasrule

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    You seem to be thinking that d[2lnx]/dx = 2lnx. It's not. It's equal to 2/x.
     
  13. Jan 3, 2010 #12

    Mentallic

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    No no, yungman never asserted that. What he did was correct, but only for finding the turning point of [itex]y=2lnx-xln2[/itex].

    It's kind of confusing with all the arrows and equals signs, so I'll clear it up:

    [tex]2lnx = xln2[/tex]

    [tex]\frac{d[2lnx]}{dx}=ln2\frac{dx}{dx}[/tex]

    [tex]\frac{2}{x}=ln2[/tex]

    [tex]x=\frac{2}{ln2}[/tex]

    But again, it doesn't help finding the solution x=4 to this problem (which btw, the OP found by an illegal move)
     
  14. Jan 3, 2010 #13
    if the problem is just as simple as solving the equation for x then just raise everything to the e.

    e^(lnx)^2=e^(ln2)^x this trivializes and is simple to solve.
     
  15. Jan 3, 2010 #14

    Mentallic

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    Yes so you have [itex]x^2=2^x[/itex] but really, this isn't as simple to solve as you claim. It's just that we are lucky enough to have nice integer solutions for x>0. If we ignored the domain on the original equation and tried to instead solve this one, you can only get a numerical approximation for the solution in [itex]-1<x<0[/itex]
     
  16. Jan 3, 2010 #15
    You loss me about "turning point"!!! What is a turning point?

    Who said that the answer is x=4? James concluded x=4 with a wrong assumption that [tex]\frac{lnx}{ln2}=ln(x-2)[/tex].

    I conclude [tex]x=\frac{2}{ln2}[/tex].....which is a constant number.
     
  17. Jan 3, 2010 #16
    Yes I tried that and didn't go no where!!!:rofl:
     
  18. Jan 4, 2010 #17

    Mentallic

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    You know about derivatives but not turning points?

    But the answer is x=4. Just try it (and btw, 4 is a constant number too :tongue:)

    [tex]xln2=2lnx[/tex]

    [tex]4ln2=2ln4[/tex]

    [tex]RHS=2ln(2^2)=4ln2=LHS[/tex]

    So x=4 is a solution. And x=2 is too, but not 2/ln2.
     
  19. Jan 4, 2010 #18
    I self study most of my calculus!!! Things sounds common to you might not be to me!!!! The only other class I took was ODE and I have people laughing at me on some stuff too!!:tongue2: But I was the first in class for the semester.

    Do you mean the point that the function turn from a convex to a concave? I remember I read something about it a few years ago. But still never heard of turning point!!!

    I still don't get why my method don't produce the answer. I know I got that by taking the derivative on both side, I have seen work problems using this method. Do I have to keep bitting my nails and wait until James come up with the answer??
     
  20. Jan 4, 2010 #19

    vela

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    It's because [tex]f(x)=0[/tex] and [tex]f'(x)=0[/tex] generally have different solutions. When you differentiated, you got a new equation, and its solution is not a solution to the original equation.
     
  21. Jan 4, 2010 #20
    Thanks, I got it.



    This is where I am at:

    [tex]xln2=2lnx\Rightarrow x=\frac{2}{ln2}x[/tex]

    [tex]\Rightarrow \frac{x}{lnx}=\frac{2}{ln2}[/tex]

    It is easy to see x=2. I still don't see x=4.
     
  22. Jan 4, 2010 #21

    HallsofIvy

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    The original equation was 2ln(x)= x ln(2). I don't see where anyone has yet pointed out that this equation cannot be solved in terms of elementary functions. It can, of course, be written as [itex]ln(x^2)= ln(2^x)[/itex] and then [itex]x^2= 2^x[/itex] and that looks like, with some manipulation, it could be solved in terms of "Lambert's W function".
     
  23. Jan 4, 2010 #22

    Mentallic

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    These were my feable attempts to do so:
    Yungman, a turning point is the point at which the gradient of the curve becomes zero. i.e. if the curve has a negative gradient but at some point it "flattens out" with 0 gradient and either becomes positive or negative again after that.

    e.g. the curve [itex]y=x(x-1)[/itex] has roots at [itex]x=0,1[/itex] but it has a turning point at [itex]x=0.5[/itex] which you can find by taking the derivative and solving for when [itex]f'(x)=0[/itex].
    This is what you did previously.
     
  24. Jan 4, 2010 #23

    Redbelly98

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    EDIT: Never mind! Clearly x>0 is required in the equation given in the OP. :redface:

    [STRIKE]I'm not quite sure why there is all this discussion of f'(x)=0. We have an equation to solve for x, as given in the OP. Discounting the x=2 solution, there are two more values of x that solve the equation. We have found x=4, but there is one more solution to find.[/STRIKE]
    [STRIKE]x2=2x[/STRIKE]​
    [STRIKE]It looks like a numerical approach is needed to find the negative solution. James889, what numerical techniques (if any) have your professor discussed?[/STRIKE]
     
    Last edited: Jan 4, 2010
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