While solving a question , I came across the following situation , we're required to find a value for $\lambda$ , from this equation :
$\lambda=(1.345)(1-e^{-\lambda})$,
I tried taking logs, but couldn't get through,
The solution on the other hand , starts off with a trial and error method. ( I have no idea how to apply trial and error method here..)
Can anyone provide me a hint ?
Hint: if the $1.345$ mattered it would be a miracle for trial-and-error to work. But it does work...
The other real solution is $1.345 + W(-1.345 \exp(-1.345))$, where $W$ is the Lambert W function.
If we consider the equation $$x=a(1-e^{-x})$$ the analytical solution is given by $$x=a+W\left(-a e^{-a}\right)$$ where $W(z)$ is Lambert function which, in the real domain, is defined if $z \geq -\frac 1e$. This function is defined by $z=W(z)e^{W(z)}$ (Euler and Lambert worked together).
You will probably learn quite soon about this marvelous function and I am sure that you will enjoy it.
Applied to the case $a=1.345$, this gives $x\approx0.625262$.
If you cannot use this function, the only solution would be a numerical maethod such as Newton which, starting at a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Writing $$f(x)=x-a(1-e^{-x})$$ $$f'(x)=1-a e^{-x}$$ which leads to the iterative scheme $$x_{n+1}=\frac{a \left(e^{x_n}-x_n-1\right)}{e^{x_n}-a}$$
For sure, there is a trivial solution $x=0$; the derivative cancels at $x=\log(a)$ which corresponds to a minimum of the function (by the second derivative test). So, if it exists, the non-trivial solution is somewhere $> \log(a)$.
For the specific value of $a=1.345$, let us start at $x_0=1$; Newton method generates the following iterates $$x_1=0.7034892906$$ $$x_2=0.6315135757$$ $$x_3=0.6253110986$$ $$x_4=0.6252619129$$ $$x_5=0.6252619098$$ which is the solution for ten significant figures.
While I can't provide a general solution, the value of $\lambda$ is approximately $0.6252619098$, which should be sufficient for your question only. Certainly, as @robert-israel suggested, look into the Lambert-W function, which should provide the means to calculate the solution more generally.
