Solution to $x=100e^{-x/100}$?

Question

How do you solve $$x=100e^{-x/100}$$

If I use $\ln$ then $\ln(x/100)=-x/100$. How do I get from that to $x=56.7$?

Answer 1

Not solvable in a closed form using algebraic functions. It is a typical and well-known trascendental equation that has as a solution a special function called the Lambert W function.

This function is defined as a principal branch of the solution of $$xe^{x}=y\rightarrow x=W(y)$$

In your case, this is almost directly what you have. Rearrange:

$$\frac{x}{100}e^{x/100}=1$$

giving

$$x=100W(1)$$

You need to solve for it numerically, or call ProductLog in Mathematica, which implements this function.

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A numerical solution is pretty straight-forward with a calculator. Write $y=x/100$ and now you have $$y=e^{-y}$$ Start somewhere reasonable (like $y=1$) and repeatedly plug it into this expression. You get a reasonable approximation in a few iterations: $1,0.3678,0.6922,0.5004,0.6062,0.5453,0.5796,0.5601,0.5711,0.5648,0.5684,0.5664,0.5675,\ldots$.

Answer 2

As the function $x\to x+\log x$ is increasing and onto, there is a unique solution to the equation $$ \log x = -x. $$

This solution has no expression in terms of elementary functions.

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The solution of your problem is obtained multiplying by 100.

Answer 3

If you want to solve numerically $$f(y)=y-e^{-y}=0$$ you can use very fast numerical methods.

For example, Newton iterations will write $$y_{new}=\frac{y+1}{e^{y}+1}$$ Starting at $y=1$ will provide the following successive iterates : $0.537883$, $0.566987$, $0.567143$ which is the solution.

As another example, Halley iterations will write $$y_{new}=\frac{2 y+2 e^y (y+2)+e^{2 y} (y (y+2)+2)+3}{2 \left(e^y+1\right)^3}$$ Starting at $y=1$ will provide the following successive iterates : $0.566599$, $0.567143$ which is the solution.