How to solve exponential equations like $2^x+x=5$?

Question

I tried the following:

Let $y=5-x$. Then, $2^{5-y}=y \implies y \cdot 2^y=32$

Taking the log of both sides yields $$\log_2 y + y = 5$$

And that's where I'm stuck.

Answer 1

Since $\frac{\mathrm{d}}{\mathrm{d}x}\left(2^x+x\right)=\log(2)\,2^x+1\gt0$, or if it is assumed that $2^x$ is monotonically increasing, we have that $2^x+x$ is monotonically increasing for all $x\in\mathbb{R}$. Since $2^1+1=3$ and $2^2+2=6$, we see that the $x$ which satisfies $2^x+x=5$ is between $1$ and $2$. However, there is no solution in terms of elementary functions.

However, if you are willing to use special functions like Lambert W: $$\newcommand{\W}{\operatorname{W}} 2^x+x=5\\ e^{x\log(2)}=5-x\\ 32\log(2)\,e^{-(5-x)\log(2)}=(5-x)\log(2)\\ 32\log(2)=(5-x)\log(2)e^{(5-x)\log(2)}\\ \W(32\log(2))=(5-x)\log(2)\\ x=5-\frac{\W(32\log(2))}{\log(2)} $$ Approximately, $x\doteq1.7156207332755861694$.

Answer 2

You are stuck for good reason. That equation does not have an algebraic solution. SInce $2^x + x$ is an increasing function of $x$ there will be a unique value of $x$ that makes it $5$. You can calculate it numerically as accurately as you like.

Answer 3

NOTE: The inverse of $f\left(x\right)=xe^x,f:\mathbb{R}\cap\left[-1,\infty\right)\rightarrow\mathbb{R}\cap\left[-\frac{1}{e},\infty\right)$ is $W\left(x\right)$, which is what is called the product logarithm function

Do not expect $x$ to have any kind of closed-form solution in terms of elementary functions, as $W\left(x\right)$ is not an elementary function itself. In fact, the inverse of $g\left(x\right)=x+2^x,g:\mathbb{R}\rightarrow\mathbb{R}$ is $g^{-1}\left(x\right)=x-\frac{W\left(2^x\log 2\right)}{\log 2}$.

So in your case, $x+2^x=5$, we have infinitely many solutions, $x\in\left\{5-\frac{W_{n}\left(32\log 2\right)}{2}:n\in\mathbb{Z}\right\}$, where $W_{n}\left(x\right)$ is the analytic continuation of the product logarithm function. And as someone below me noted, there is only one branch of the product log function for nonnegative real numbers so you have only one real solution. All of the other solutions are complex numbers.

Answer 4

Without using Lambert function (which I am in love with !), you can approximate the function building around $x=2$ (which would be the solution of $2^x+x=6$) the simplest $[1,1]$ Padé approximant of $$f(x)=2^x+x-5$$ This would give $$2^x+x-5 \approx \frac{1+\frac{ 1+8 \log (2)+14 \log ^2(2)}{1+4 \log (2)}(x-2)}{1-\frac{2 \log ^2(2)}{1+4 \log (2)}(x-2)}$$ Then $$x\approx \frac{1+28 \log ^2(2)+12 \log (2)}{1+14 \log ^2(2)+8 \log (2)}\approx 1.71574$$ Better would be the result using instead the $[1,n]$ approximant and get an approximation still at the price of a linear equation. For example, the next one would give $$x\approx \frac{3+292 \log ^3(2)+222 \log ^2(2)+48 \log (2)}{3+146 \log ^3(2)+132 \log ^2(2)+36 \log (2)}\approx 1.71560$$