# How to solve this implicit equation? (use trial and error)

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 ?

## 1 Answer

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 $$x=W(e^b)$$.

• 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)
– User
Jun 3, 2020 at 4:20
• @Soheil0098: Then we use $x=\frac{\ln 10}{2t}$ and get the same with $b=\ln\Big(\frac{a}{2}\ln 10\Big)-0.4\ln 10$. Jun 3, 2020 at 6:52
• 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?
– User
Jun 3, 2020 at 11:18
• 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?
– User
Jun 3, 2020 at 11:24
• @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.) Jun 3, 2020 at 11:24