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Algebra: Laws of Logs

  1. Aug 2, 2009 #1
    I need to solve for t and it's slipped my mind how to manipulate this.

    [tex]\frac{1}{4}=te^{-8t}[/tex]

    to

    [tex]ln(1/4)=lnt-8t[/tex]


    I understand the laws of logs (I think), but I still can't seem to isolate the t.
     
  2. jcsd
  3. Aug 2, 2009 #2
  4. Aug 2, 2009 #3

    jgens

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    Gold Member

    Well, you could also prove that there isn't any value of t that satifies that equation. It's pretty simple using derivatives . . .

    Proof: Consider the function f defined such that f(t) = exp(8t) - 4t. Clearly, if for some value t0 we have that f(t0) = 0 then 4-1 = t0exp(-8t0). Now, we need only prove that there is no value t0 with this property.

    From the definition of f(t) we certainly have that f '(t) = 8exp(8t) - 4. Since f '(t) is defined for all real numbers, the only critical points can occur when f '(t0) = 8exp(8t0) - 4 = 0 => t0 = - ln(2)/8. Since f '(t) < f '(t0 for t < t0 and f '(t) > f '(t0 for t > t0, we know that f(t0) is an absolute minimum. However, because f(- ln(2)/8) = 2-1(1 + ln(2)) > 0, there is no value t0 such that 4-1 = t0exp(-8t0). Q.E.D.
     
  5. Aug 3, 2009 #4
    I think I can express the derivative in terms of the inverse function in terms of itself f(y)=t

    [tex]f(te^{-8t})=t[/tex]

    Differentiating both sides:

    [tex]f'(te^{-8t})\left(e^{-8t}-8te^{-8t}\right)=1[/tex]

    [tex]f'(te^{-8t})=\frac{1}{e^{-8t}-8te^{-8t}}[/tex]

    [tex]f'(y)=\frac{1}{\frac{1}{f(y)}y-8y}[/tex]

    Which might have some potential for numeric algorithms or evaluating the power series.

    Lets try a big more. At the point you are are interested in y=1/4

    [tex]f'(1/4)=\frac{1}{\frac{1}{t}(1/4)-8(1/4)}[/tex]

    [tex]f'(1/4)=\frac{1}{\frac{1}{t}(1/4)-2}[/tex]

    [tex]\frac{1}{t}(1/4)-2=\frac{1}{f'(1/4)}[/tex]

    [tex]t=\frac{1}{4}\frac{1}{\frac{1}{f'(1/4)}+2}[/tex]

    Not sure if this helps at all.
     
    Last edited: Aug 3, 2009
  6. Aug 3, 2009 #5
  7. Aug 3, 2009 #6

    Gib Z

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    Homework Helper



    That's somewhat hard to read, I'm not sure what exactly you did.

    I was thinking of something like : f'(x) = 0 at x=1/8 only , f''( 1/8) < 0, hence f( 1/8) is a max. Evaluating, it is clearly less than 1/4.
     
  8. Aug 3, 2009 #7

    jgens

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    Gold Member

    Well, here's a revised edition of the proof. I'm new to LaTeX so it doesn't look very neat . . .

    Statement: [tex](\forall t)(t \in \mathbb {R})\left (\frac{t}{e^{8t}} \neq \frac{1}{4} \right )[/tex]

    Proof: Consider the function [tex]f[/tex] defined such that [tex]f(t) = e^{8t} - 4t[/tex]. From the definition of [tex]f[/tex], if for some [tex]t = t_{0}[/tex] we have that [tex]f(t_{0}) = 0[/tex], then

    [tex]\frac{t_{0}}{e^{8t_{0}}} = \frac{1}{4}[/tex]​

    Now, consider the first derivative of [tex]f[/tex] so that [tex]f '(t) = 8e^{8t} - 4[/tex]. Since [tex]f '[/tex] is defined [tex]\forall t \in \mathbb {R}[/tex], the only critical points can occur when [tex]f '(t_{0}) = 0[/tex]. Solving for the critical points, we find that [tex]e^{8t_{0}} = 1/2[/tex]. Since [tex]e^{8t} > 0[/tex], we don't lose any solutions when taking the logarithm base [tex]e[/tex] of the previous functions, therefore [tex]t_{0} = - ln(2)/8[/tex]. Using a similar method, we can show that [tex]f '(t) < 0[/tex] if [tex]t < t_0[/tex] and similarly [tex]f '(t) > 0[/tex] if [tex]t > t_0[/tex]. This proves that [tex]f(t_0)[/tex] is an absolute minimum. However,

    [tex]f(t_0) = e^{-ln(2)} + \frac{ln(2)}{2} = \frac{1}{2} + \frac{ln(2)}{2} = \frac{1}{2}(1 + ln(2)) > 0[/tex]​

    which proves that there is no value of [tex]t[/tex] such that [tex]f(t) = 0[/tex].

    So, I don't think that there's any need to invoke the lambert W function or power series, just really simple calculus. :)
     
  9. Aug 3, 2009 #8
    Complex solutions exist. The formula is not elementary, but can be written using the Lambert W function. t = -W(-2)/8 . One value is about -0.02160200035 - 0.2092108018 i
     
  10. Aug 4, 2009 #9
    It should be [tex]f(t) = e^{-8t} - 4t[/tex]. which pretty much invalidates the rest of your post.
     
  11. Aug 4, 2009 #10
    Cool, thanks. :) I worked this out on wikipedia and got the same thing.
     
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