Solving $\dfrac{x}{1-e^{-x}} = 3.034$

Question

I am stuck in this question in finding the value of $x$ :-

$$\frac{x}{1-e^{-x}} = 3.034$$

I have tried this question using log on both sides.

Like this:-

$$\log x - \log (1-e^{-x}) = \log (3.034)$$

$$\frac{1}{x\:} - \frac{1}{1-e^{-x}} =\log(3.034)$$

Given answer:- $x= 2.86$

But it turned out bad after taking l.c.m

I don't know how to solve after this..

Please help me & correct me if I have done something wrong calculations.

Answer 1

That's a form of transcendental equation. So it has no elementary solution, you can only approximate it.

In general, if you see $x$ both inside and outside a trig function, log, or exponential, the equation is transcendental and you have to use numerical techniques to get close to an answer.

The simplest practical way is to rewrite it as two functions of $x$ that are equal, and graph them both, then find the point of intersection.

Answer 2

The analytic solution to:

$$\frac{x}{1 - e^{-x}} = a$$

is

$$x = W\left(-a e^{-a}\right)+a$$

where $W$ is the product log function or Lambert's $W$ function.

Evaluated with $a = 3.034$ gives $$x = 2.8603.$$

Answer 3

As @David G. Stork answered, the only explicit solution involves Lambert function.

Assuming that you cannot use it, you can have quite good approximations for the non trivial zero of function $$f(x)=x-a(1-e^{-x})\qquad \text{with} \qquad a >0$$ for which $$f'(x)=1-a e^{-x} \qquad \text{and} \qquad f''(x)=a e^{-x} \quad >0 \quad \forall x$$

This firstderivative cancel at $$x_*=\log(a) \implies f(x_*)= -a+\log (a)+1 \le 0$$ So, around $x=x_*$, make a simple series expansion $$f(x)=f(x_*)+\frac 12 f''(x_*)(x-x_*)^2+ O\left( (x-x_*)^3\right)$$ which gives $$x \sim x_*+\sqrt{-2 \frac{f(x_*)}{f''(x_*)}}$$ that is to say $$x_0=\log(a)+ \sqrt{2(a-\log (a)-1)}$$ If $a=3.034$, this gives $x_0=2.46938$.

Now, perform a single iteration of Newton method; it would give $x_1=2.88357$ which is not too bad.