$$908.80 = 250(1+\frac{0.24}{m})^{6m}$$

I get stuck here because one of the m's is part of the ln:

$$ln908.8 = m(6ln250 + 6ln250\frac{0.24}{m})$$

Not sure how to pull it out. I've tried various things but keep getting the wrong answer; I know the final answer is supposed to be m=0.99. Any help?

  • $\begingroup$ Recall, $ln(\frac{a}{b})=ln(a)-ln(b)$ $\endgroup$ – 高田航 Mar 30 '17 at 3:24
  • $\begingroup$ You're not going to be able to find a final answer to this algebraically. $\endgroup$ – Kaynex Mar 30 '17 at 3:25
  • 1
    $\begingroup$ Your second line doesn't follow from your first line. Did you try to distribute log over a sum? $\endgroup$ – Brian Tung Mar 30 '17 at 3:32
  • $\begingroup$ $$6mln250(1+\frac{0.24}{m})$$, then I distributed it, then factored out the m $\endgroup$ – minicat Mar 30 '17 at 4:35

Without the W function you are in for a numeric solution. I like fixed point iteration, so would take logs $$908.80 = 250(1+\frac{0.24}{m})^{6m}\\\log 908.80=\log 250+6m\log(1+\frac {0.24}m)\\m=\frac{\log 908.80-\log 250}{6 \log(1+\frac{0.24}m)}$$ and iterate $$m_{i+1}=\frac{\log 908.80-\log 250}{6 \log(1+\frac{0.24}{m_i})}$$ to convergence at about $m=0.999968$

  • $\begingroup$ Thank you, this is helpful. $\endgroup$ – minicat Mar 30 '17 at 4:31

You will need the Lambert-$W$ function for this. There is no solution in terms of elementary functions.

EDIT: Rewrite the equation as


Letting $1/m=x$, we have


which is of the form $p^x=ax+b$ and whose solution is in general given by

$$x=\frac{-W\left(-\frac{\ln p}{a}\,p^{-b/a}\right)}{\ln p}-\frac{b}a$$

and making the approipriate substitutions yields $m$ in terms of the Lambert-$W$ function.

  • $\begingroup$ I am not sure that Lambert function could be used for this problem. $\endgroup$ – Claude Leibovici Mar 30 '17 at 4:57
  • $\begingroup$ @ClaudeLeibovici I have clarified how to use it for this problem; see the edit. $\endgroup$ – Fimpellizieri Mar 30 '17 at 5:11
  • $\begingroup$ I apologize for my stupidity ! $\endgroup$ – Claude Leibovici Mar 30 '17 at 5:28
  • $\begingroup$ That wasn't stupid, it was just wrong. No need to apologize; we're all wrong at times. We all make mistakes. $\endgroup$ – Fimpellizieri Mar 30 '17 at 6:02

As Ross Millikan answered, a numerical method is required. So, considering that me need to find the zero of equation $$f(m)=\log (250)+6m\log\left(1+\frac {0.24}m\right)-\log( 908.80)$$ that is to say $$f(m)=m\log\left(1+\frac {0.24}m\right)-\frac16 \log\left(\frac {908.8}{250}\right)$$ let me be very lazy and start Newton iterations using $m_0=0.5$. The successive iterates will be $$\left( \begin{array}{cc} n & m_n \\ 0 & 0.50000000 \\ 1 & 0.78190010 \\ 2 & 0.95860749 \\ 3 & 0.99848163 \\ 4 & 0.99996602 \\ 5 & 0.99996794 \end{array} \right)$$



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