Any advice on reducing $(x + 1.07142)e^{-14x} = 33\times 10^{-10}$

Question

I am working on a research project and came unstuck when I got to this point. I am generating a numerical solution for some variable $x$ but cannot seem to figure out how to reduce the expression below. The equation is

$$(x + 1.07142)e^{-14x} = 33\times 10^{-10}$$

I know the result ($x = 1.4615$..), but cant seem to figure out the solution to get that result. Any suggestions on how to go about this will be deeply appreciated.

Thank you.

Answer 1

There are two roots for $$ f(x) = \left( x+1.07142)e^{−14x}-33×10^{−10} \right), $$ $$ x_{1} = -1.07142, \quad x_{2} = 1.461332134293102. $$ There is no closed form solution.

The plot below of $$ \log f(x) $$ shows the location of these roots. Outside the interval bounded by the roots the value of the function is negative, and the logarithm is complex

Answer 2

There are many one dimensional root finding algorithms and any numerical analysis text will discuss them. I like the discussion in Numerical Recipes with obsolete versions free on line. Once you note that the root is somewhere between $1$ and $2$ any of the root finders should have an easy time. If I just have one problem to do I often look for fixed point iteration. If you can write your equation in the form $f(x)=x$ with $f(x)$ slowly varying it will converge rapidly. Here I would write $$(x + 1.07142)e^{-14x} = 33\times 10^{-10}\\ x=-\frac 1{14}\ln\left(\frac{33\times 10^{-10}}{x + 1.07142}\right)$$
start with $x_0=1.5$ and iterate to convergence. Writing this in a spreadsheet is easy with copy down for the iteration. It converges to machine accuracy in six iterations.

Answer 3

As mathematician commented, the solution of equation $$(a+x) e^{-b x}=c$$ can be expressed in terms of Lambert function (which is not an elementary function).

Its expression is $$x=-a-\frac{1}{b}\,W\left(-b\, c\, e^{-a\, b}\right)$$ Tha Wikipedia page shows on many examples the manipulations to be done.

Applied to your case, this would give for the negative root $$x_1=-1.07142-\frac 1{14} W(-1.41344\times 10^{-14})$$ and for the positive root $$x_2=-1.07142-\frac 1{14} W(-1,-1.41344\times 10^{-14})$$ The Wikipedia page gives series expansion for the evaluation of these. For the first one, since $W(\epsilon)\approx \epsilon$, then $x_1\approx -1.07142$.

For the second one, it is slightly more complex but you can use

$$W(-1,x)={L_1}-{L_2}+\frac{{L_2}}{{L_1}}+\frac{({L_2}-2) {L_2}}{2 {L_1}^2}+\cdots$$ where $L_1=\log(-x)$ and $L_2=\log(-L_1)$. Here $L_1=-31.8902$, $L_2=3.4623$ and then $W(-1,-1.41344\times 10^{-14})\approx -35.4585$. This makes as a result $x_2\approx 1.46133$.

As Ross Millikan answered and commented, Newton method would do a great job. Starting using $x_0=1.5$, the successive iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.500000000 \\ 1 & 1.449123485 \\ 2 & 1.460372939 \\ 3 & 1.461325908 \\ 4 & 1.461332134 \end{array} \right)$$ It could have been faster writing the equation as $$\log(x + 1.07142)-14x = \log(33\times 10^{-10})$$ Starting using $x_0=1.5$, the successive iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.500000000 \\ 1 & 1.461340525 \\ 2 & 1.461332134 \end{array} \right)$$ Sooner or later, you will learn that any equation which can write or rewrite $$A+Bx+C\log(D+Ex)=0$$ has real and/or complex solution(s) in terms of Lambert function .