Solving the following equation: $\ln(x)=\frac{x}{1+x}$

Question

I am stuck with solving this equation for $x$:

$$\ln(x)=\frac{x}{1+x}$$

Making exponential both sides I get:

$$x=\exp\left(\frac{x}{1+x}\right)$$

Then I tried to make a change of variable $x=\frac{1}{z}$ and use omega function. But I reached nothing.

Any help please?

Answer 1

I do not think there is a closed form solution. Hence, I will use the Newton-Raphson Method.

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One may prove that there exists only one real root.

The process is as follows:

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \tag{1}$$

We choose an initial starting point $x_0=1$, a reasonable estimate of the solution.

We use the functions:

$$f(x)=\ln{x}-\frac{x}{x+1}$$

And find its derivative:

$$f'(x)=\frac{x^2+x+1}{x(x+1)^2}$$

Hence, we obtain the following, after substituting into $(1)$:

$$x_{n+1}=x_n-\frac{\ln(x_n)-\frac{x_n}{x_n+1}}{\left(\frac{{x_n}^2+x_n+1}{x_n(x_n+1)^2}\right)}=x_n-\frac{x_n(x_n+1)((x_n+1)\ln(x_n)-x_n)}{{x_n}^2+x_n+1} \tag{2}$$

Applying this recursively gets us closer and closer to the solution:

$$\begin{array}{c|c}n&x_n\\\hline0&1\\1&1.66667\\2&1.91521\\3&1.93287\\4&1.93295\\5&1.93295\end{array}$$

As the iterations $n \to \infty$, $x_n \to x$, the solution to the equation. Therefore:

$$x \approx 1.932947554905753$$

You can implement this method using a spreadsheet, or by a more sophisticated software such as MATLAB.

Answer 2

Hint. The function $f(x)=\ln(x)-\frac{x}{1+x}$ is strictly increasing for $x>0$: $$f'(x)=\frac{1+x+x^2}{x(1+x)^2}>0.$$ Moreover $f$ is continuous, $f(1)=-1/2<0$ and $f(2)=\ln(2)-2/3>0$. So the equation $f(x)=0$ has a unique solution which belongs to the interval $(1,2)$. Finding the exact value of such a solution in a closed form is not trivial. However you can approximate it by using the Bisection Method or the Newton's method.