Can the equation $2^x + x = 11$ be solved without graphing?

Question

$2^{x}+x = 11$

Well this problem is easy to solve just by looking at its graph, and we find the answer is $x = 3$, but I want a way of solving it rather than just looking at it to find the solution. I realized it was harder than it seemed or I am missing something obvious. Trying to use exponent and logarithmic rules all ended up with a dead end, and the graphing solution is unsatisfactory to me.

Answer 1

Rearrange the equation to get $2^x=11-x$. Note that the left side is an increasing function and the right side is a decreasing function. Hence, they will intersect at most one point. There isn't really any way to actually solve the equation except trial and error(feel free to correct me on this) but once you find a solution($x=3$), you will know that it is the only one.

Edit: From the comments it seems that you could use the Lambert W Function, but it may be beyond the scope of this discussion.

Answer 2

The answer is yes, but the methods are beyond anything in "algebra-precalculus."

The solution involves the "Lambert-W" function. So, what is the Lambert W?

$xe^x = z \iff x = W(z)$

So, what you need to do is get your expression into the form of $ue^u = z$

$2^x = 11-x$
$(2^{x-11})2^{11} = 11-x$
$2^{11} = (11-x)2^{11-x}$
$2^{11} = (11-x)e^{(\ln2)(11-x)}$
$(\ln2)(2^{11}) = (\ln2)(11-x)e^{(\ln2)(11-x)}$

Okay, we have it in the right form. We could say $u = (\ln2)(11-x)$ and $(\ln 2)(2^{11}) = ue^u$ to make it more obvious.

$u = W((\ln 2)(2^{11}))$
$x = 11-\frac {W((\ln 2)(2^{11}))}{\ln2}$

And then I would need a claculator like wolfram-alpha to indeed see that $\frac {W((\ln 2)(2^{11}))}{\ln2} = 8$

https://www.wolframalpha.com/input?i=lambert+W%282%5E11ln2%29%2Fln2

Answer 3

Sometime an equation of this sort can be solved with the method of successive approximations. For instance, consider the equation

$$2^x + x = 12\ \ (*)$$

We do not have the obvious solution $x=3$ like when RHS $= 11$. So what we do is

1. Transform it into an equation amenable to successive approximations

$$x= \log_2 (12 -x)\ \ (**) $$

2. Start with an initial seed $x_0$, and construct the sequence

$$x_{n+1} = \log_2(12 - x_n)$$

The sequence ( for most initial seeds $x_0>0$) converge to the unique positive solution of (*). For instance, start with $x_0$ = 3$. Then we get the sequence

$$3, 3.1699\ldots, 3.1424\ldots, 3.1469\ldots, 3.14618\ldots, 3.1463\ldots, 3.14628$$

and the sequence appears to stabilize its first $5$ decimals. This would be the idea.

Now, let's see what would happen if we dealt directly with the equation (*).

Again take $x_0 = 3$, and $x_{n+1}= 12 - 2^x$. It's pretty clear that there is no way this sequence will approximate the solution of (*), since its terms are all integers. Moreover, one checks that the sequence oscilates widely.

Answer 4

Without special function.

Consider that you look fot the zero of the strictly increasing function $$f(x)=2^x+x -a$$ where $a$ is a huge number.

Taking into account the fact that $2^x+x >2^x$, we then have a first estimate given by $$x_0=\log_2(a)$$ and, because of the above remark, the solution of $f(x)=0$ will we smaller than $x_0$.

To make the function closer to linearity, now swith the problem to $$g(x)=\log(2^x)-\log(a-x)=x\log(2)-\log(a-x)$$ and make one iteration of Newton method around $x_0$. This will give $$x_1=\frac{\frac{x_0}{a-x_0}+\log (a-x_0)}{\frac{1}{a-x_0}+\log (2)}$$

Trying for $a=12345$ we have (converted to decimals) $$x_0=\color{red}{13.59}1639216030144 \qquad \text{and} \qquad x_1=\color{red}{13.5900501437161}66$$ while the exact solution is $x=\color{red}{13.590050143716154}$

Answer 5

1)

One simple way to try to solve your equation without graphing is to guess and test possible solutions.
But the most general way is to solve it numerically.
$\ $

2)

Now I want to write why your trials had to fail.

$$2^x+x=11$$

Your equation consists only of elementary operations/functions.

$$2^x+x-11=0$$

Solving the equation by rearranging it by applying only elementary operations/functions means to find elementary (partial) inverses of the elementary function on the left-hand side of the equation.

The elementary functions are generated by applying finite numbers of $\exp$, $\ln$ and/or algebraic functions.

Your equation isn't an algebraic equation (polynomial equation). So the solution formulas/methods for solving polynomial equations cannot be applied here.

Your equation is

$$e^{\ln(2)\ x}+x-11=0.$$

The elementary function on the left-hand side of this equation is a binary algebraic function in dependence of $e^{\ln(2)\ x}$ and $x$. Because $e^{\ln(2)\ x}$ and $x$ are algebraically independent, the function expression on the left-hand side of this equation cannot be simplified to the function term of a unary algebraic function in dependence of only one elementary function (a function composition).

A theorem of Ritt (1925) states that the elementary functions with elementary (partial) inverses are compositions of elementary functions. Therefore your elementary function doesn't have non-constant elementary (partial) inverses, and your equation cannot be solved by simply applying elementary functions you can read from the equation. Therefore your trials to use exponent and logarithmic rules all ended up with a dead end.
It's an open question if the equation has solutions except $3$ that are elementary numbers.

But there is a nonelementary special function Lambert W that gives the partial inverses of a binary algebraic function in dependence of two elementary functions: $xe^x=y\implies x=W(y)$.

3)

Your equation can be solved by Lambert W, as other answers of this discussion state.