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Thread: Logarithmic Equation

  1. #1

    Logarithmic Equation

    Can someone show me how to solve this?

    ln(x-2)+lnx=0

  2. #2
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    Quote Originally Posted by harrypotterkid View Post
    Can someone show me how to solve this?
    ln(x-2)+lnx=0
    Can you solve [TEX]x(x-2)=1~?[/TEX]
    Why does that work?
    “A professor is someone who talks in someone else’s sleep”
    W.H. Auden

  3. #3
    yes. so do I just use the distributive property?

  4. #4
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    Quote Originally Posted by harrypotterkid View Post
    yes. so do I just use the distributive property?
    Solve [TEX]x^2-2x-1=0[/TEX]
    “A professor is someone who talks in someone else’s sleep”
    W.H. Auden

  5. #5
    alright. 2.414 and -0.414

  6. #6
    Is the answer to my origional problem 3.183?

  7. #7
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    harrypotterkid, when solving problems with logs, no matter what the base, try and condense the individual log functions into one log function so you get the following structure:

    [tex]\displaystyle log_b[f(x)]=a[/tex] where "a" is some constant.

    Then you can use the property that [tex]\displaystyle b^a=f(x)[/tex] and then you can solve for the variable, x.

    So for this problem of yours, remember [tex]\displaystyle log_b(u)+log_b(v)=log_b(u\cdot v)[/tex] thus,

    [tex]\displaystyle ln(x-2)+ln(x)=0 \longrightarrow ln[(x-2)\cdot x]=0[/tex]

    Then, [tex]\displaystyle e^0 = (x-2)x[/tex]. Thus,

    [tex]\displaystyle x^2-2x=1 \longrightarrow x^2-2x-1=0[/tex]

    Now, when you solved this you got [tex]\displaystyle 1 \pm \sqrt{2}[/tex], which is correct as far as what one gets using the quadratic fucntion, however, one must check each solution to make sure they both satisfy the original problem and in this instance, [tex]\displaystyle 1-\sqrt{2}[/tex] does not as you cannot take the log of 0 or, in this case, a negative number. So [tex]\displaystyle 1-\sqrt{2}[/tex] is an extraneous solution and the only solution is [tex]\displaystyle 1+\sqrt{2}[/tex]
    "There are 10 types of people in this world - those who understand binary and those who don't."

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