1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help with logarithms
  1. Logarithm help (Replies: 5)

Loading...