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I want to solve this implicit equation and find $f$ When $Re$ is constant: $$\frac{1}{\sqrt{f}}=2\log({Re.\sqrt{f})}- 0.8$$

I tried to make the equation simple By using: $\sqrt{f}=t>0,Re=a$: $$\frac{1}{t}=2\log({at})-0.8$$

$$0.8t=2t\log(at)-1$$

I can not find $t$ in this eqution. Should I draw graphs of $\frac{1}{t},2\log({at})-0.8$ and find intersection points for given "a" as question said with "trial and error method". But in general I don't like use this method for math questions. So is it possible to solve equation in other way ?

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1 Answer 1

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It can be solved in terms of the Lambert $W$ function: $$\frac{1}{2t}+\log\frac{1}{2t}=b:=\log\frac{a}{2}-0.4$$ or, for $x=1/(2t)$, we have $xe^x=e^b$, thus $x=W(e^b)$.

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  • $\begingroup$ Thanks, I have a question you said for x=1/(2t) we have $xe^x=e^b$ but it is true if the base of $log(\frac{1}{2t})$ is $e$ but here the base is 10 I think (It is a formula of physics and it doesn't say what is the logarithm base so I think It is 10) $\endgroup$
    – User
    Jun 3, 2020 at 4:20
  • $\begingroup$ @Soheil0098: Then we use $x=\frac{\ln 10}{2t}$ and get the same with $b=\ln\Big(\frac{a}{2}\ln 10\Big)-0.4\ln 10$. $\endgroup$
    – metamorphy
    Jun 3, 2020 at 6:52
  • $\begingroup$ Here:$$\frac{1}{2t}+\log\frac{1}{2t}=\log\frac{a}{2}-0.4$$ $$=\frac{1}{2t}+\frac{ln(\frac{1}{2t})}{ln10}=\frac{ln(\frac{a}{2})}{ln10}-0.4$$ $$=\frac{ln10}{2t}+ln(\frac{1}{2t})=ln(\frac{a}{2})-0.4×ln10 (consider\frac{ln10}{2t}=x)$$ $$\frac{xe^x}{ln10}=\frac{\frac{a}{2}}{10^{0.4}}$$ $$xe^x=\frac{a×ln10}{2×10^{0.4}}$$ Is this True? $\endgroup$
    – User
    Jun 3, 2020 at 11:18
  • $\begingroup$ I am not familiar with Lambert W function. If it is elementary to solve with Lambert Can you please show me how it should be solved? $\endgroup$
    – User
    Jun 3, 2020 at 11:24
  • $\begingroup$ @Soheil0098: Yes, your derivation is correct. The Lambert function is not elementary, and there's no elementary solution. (BTW, for future: use \ln, not just ln, and similarly \log, \sin, etc.) $\endgroup$
    – metamorphy
    Jun 3, 2020 at 11:24

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