How to solve equation with exponents multiplied with a polynomial?

Question

I don't know how to tackle equation with exponents multiplied with a polynomial in general.

For example:

$100 =2^x(1+x)$

Currently, I ignore the polynomial part since it grows much slower. I solve for $100=2^x$ and let algorithm to test out numbers around the range.

Answer 1

You can solve this with the Lambert W function: Write you equation as $$100 = 2^x(1+x)=\frac{1}{2}2^{1+x}(1+x),$$ with $y=1+x$ you get $$200 = y2^y=y e^{y\ln 2} = \frac{1}{\ln 2} (y \ln 2) e^{y \ln 2} .$$ Now substitute $z=y \ln 2$ and use the definition of $W$ to solve for $z$ $$200 \ln 2 = z e^z \quad \Rightarrow z = W(200\ln 2) \quad \Rightarrow y = \frac{W(200\ln 2)}{\ln2}$$ So finally

$$x = \frac{W(200\ln 2)}{\ln2} -1 \approx 4.2512070962222326$$

As an alternative you can use a simple iteration scheme from the equation $$x_{n+1} = \ln \left(\frac{100}{1+x_n}\right) / \ln 2 $$ With your starting value $x_0=\log_2(100) \approx 6.6439$ you get the next iterates $3.7094, 4.4083, 4.2086, 4.2630, \dots$

Answer 2

As a rule of thumb, equation that mix transcendental functions and polynomials have no closed-form solutions and you need to resort to numerical methods for the resolution of equations (such as Newton's).

In this particular case, you can reduce to the form

$$ze^z=c,$$ which was studied by Lambert. He defined a special function, which precisely gives the solution of this equation.