How to solve an equation like $2{^x} + x = 2 $?

Question

I encountered an equation similar to this in an old math exam.

$2{^x} + x = 2 $

I couldn't figure it out and ended up with a mess of logarithmic functions. The answer sheet indicated it should be solved using a graphing calculator, which did work. When I asked the teacher how to do it without a calculator he said that he didn't have time to explain it because it was too complicated.

I'd like to know how these kinds of equations could be solved, because I'm at a loss here. I haven't been able to find anything online either.

Answer 1

This answer can not be found without a calculator. There is no exact answer to this problem.

Numbers can be divided into two groups - Algebraic Numbers and Transcendental Numbers. Algebraic numbers are ones which are the solution to polynomials with integer (or some definitions say rational) coefficients. The solution to your equation can not be expressed as such.

Answer 2

Notice, you've to use the Product log function:

$$a^x+x=a\Longleftrightarrow x=\frac{a\ln(a)-\text{W}_n\left(a^a\ln(a)\right)}{\ln(a)}$$

Where: $\text{W}_k(z)$ is the analytic continuation of the product log function and $n\in\mathbb{Z}$

The real solution is given by $(n=1)$:

$$a^x+x=a\Longleftrightarrow x=\frac{a\ln(a)-\text{W}\left(a^a\ln(a)\right)}{\ln(a)}$$

In your example $a=2$:

$$2^x+x=2\Longleftrightarrow x=\frac{2\ln(2)-\text{W}\left(4\ln(2)\right)}{\ln(2)}\approx0.5430004$$

Answer 3

$$2^x=2-x$$
And it is the only real solution, because $2^x$ is increasing, but $x-2$ is decreasing.

Answer 4

There is no closed form solution to this. Note that $2^x+x$ is a monotone increasing function. Hence, there exists a unique real solution. An approximate solution can be found using root finding methods such as Newton Raphson or the bisection method. Note that if $f(x) = 2^x+x-2$, we then have that $$f(0) = -1 <0 < 1 = f(1)$$ Hence, the unique root is in the interval $[0,1]$. Approximate solution using bisection gives me $x^* \approx 0.543000$