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Dumb log question

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data
    27x+1=32x+1


    3. The attempt at a solution

    log(27x+1)=log(32x+1)

    (x+1)log(27)=(2x+1)log(3)

    2x+1=[itex]\frac{(x+1)log(27)}{log(3)}[/itex]

    [itex]\frac{2x+1}{x+1}=\frac{log(27)}{log(3)}[/itex]

    x=-2

    I'm wondering how I would have continued solving this for the exact answer if [itex]\frac{log(27)}{log(3)}[/itex] had turned out to be irrational?
     
  2. jcsd
  3. Apr 15, 2012 #2

    Curious3141

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    Homework Helper

    If log(m)/log(n) was irrational (call it p), then you'd just have solved it with algebra the usual way to get x in terms of p.

    i.e.

    (2x+1) = px + p

    x(2-p) = p-1

    x = (p-1)/(2-p)

    That would be an exact answer, but it wouldn't be a "nice number" (an integer or even a rational answer).
     
  4. Apr 15, 2012 #3

    cepheid

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    Staff Emeritus
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    Gold Member


    Saying "log(a)" and "log(b)" IS exact, even if you can't express those as rational numbers.

    xlog(a) + log(a) = 2x(log(b)) + log(b)

    x[log(a) - 2log(b)] = log(b) - log(a)

    [tex] x = \frac{\log(b) - \log(a)}{\log(a) - 2\log(b)} [/tex]

    and that would be your final answer, which is very much exact.
     
  5. Apr 15, 2012 #4
    (2x+1) = px + p

    x(2-p) = p-1

    I am probably being incredibly dense, but I don't follow what you did here.
     
  6. Apr 15, 2012 #5
    Yeah I realize that, it's like [itex]\sqrt{2}[/itex]. The problem I'm having is with the algebra.
     
  7. Apr 15, 2012 #6
    Got it, I really was being dense. Thank you.
     
  8. Apr 15, 2012 #7
    [tex]\frac{2x+1}{x+1}=\frac{a}{b}[/tex]
    [tex]2x+1=\frac{a}{b}(x+1)[/tex]
    [tex]2x+1=\frac{a}{b}x+\frac{a}{b}[/tex]
    [tex]2x-\frac{a}{b}x+1=\frac{a}{b}[/tex]
    [tex]2x-\frac{a}{b}x=\frac{a}{b}-1[/tex]
    [tex] x\left (2-\frac{a}{b}\right )=\frac{a}{b}-1[/tex]
    [tex] x=\frac{\frac{a}{b}-1}{2-\frac{a}{b}}[/tex]

    edit: nvm, you got it
     
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