# Solve for $x$: $4^x + x = 11$

I came across this question on Facebook (as a puzzler): \begin{align}\\ 2^x + x = 11\\ \end{align}

I solved this by using the following:

Since $$2^x$$ is even, therefore $$x$$ must be odd. So $$\exists k \in \Bbb{Z}$$ such that $$x = 2k + 1$$.

So $$2^x + x = 11 \implies 2^{2k + 1} + (2k + 1) = 11$$

Then $$2^{2k + 1} + 2k = 10$$, $$\implies$$ $$2^{2k} + k = 5$$ Repeating the process, $$\exists r \in \Bbb{Z}$$ such that $$k = 2r + 1$$.

So $$2^{2k} + k = 5$$ $$\implies$$ $$2^{2(2r + 1)} + (2r + 1) = 5$$

Then $$2^{4r + 2} + 2r + 1 = 5$$

So $$2^{4r + 2} + 2r = 4$$,

Then $$2^{4r + 1} + r = 2$$ Now, this means r is even. so $$r = 2m$$

$$\implies$$ $$2^{8m + 1} + 2m = 2$$, $$\implies$$ $$2^{8m} + m = 1$$.

At this point, this is only valid only if $$m = 0$$.

$$\therefore$$ $$r= 0$$ and so $$k = 1$$ and finally, $$x = 3$$ I even did a graph on Excel, and that pretty much was the answer.

Then I thought about $$3^x + x = 11$$. (Answer is $$x = 2$$, though I had a bit of a problem figuring this one out.)

Now, I came up with the next item. $$4^x + x = 11$$.

At this point, I know it isn't going to have an integer solution since if $$x = 1$$, $$4^x + x = 4 + 1 = 5 \not= 11$$

Numerically, using python, I got the value to $$1.614999 < x < 1.615999$$. However, I was hoping to prove this.

So I came up with the following (but fell short).
\begin{align} \\ 4^x + x = 11 \\ \implies 2^{2x} + x = 11\\ \end{align}

And if $$x = 2$$, $$4^x + x = 16 + 2 = 18 \not= 11$$

\begin{align}\\ 2^{2x} + x = 11 \\ \implies 2^{2x} + x - 11 = 0 \end{align}

My first thought was to try using substitution:

\begin{align} let\ u = 2^x, so\ x = \frac{\ln{u}}{\ln{2}}\\ Then\ u^2 + \frac{\ln{u}}{\ln{2}} - 11 = 0\\ \end{align}

But this is not a quadratic equation, so I can't use the quadratic formula.

So I tried using the 'find the a,b of $$(a - b)^2 = a^2 - 2ab + b^2$$

\begin{align} let\ a^2=2^{2x},\ b^2 = x,\ -2ab = -11, \implies ab = \frac{11}{2}\\ Then\ b = \frac{11}{2a},\implies b^2 = \frac{11^2}{2^2a^2} \\ But\ b^2 = x, \\ \therefore, x = \frac{11^2}{2^2a^2}\\ and\ so\ x = \frac{11^2}{2^22^{2x}} \\ or\ x2^{2x} = \frac{11^2}{2^2} [1]\\ \end{align}

At this point, things are losing sense; but for fun(and to see what happens), I came up with this mess [I feel that $$x2^{2x} = \frac{11^2}{2^2}$$ is suspiciously wrong]:

Since $$2^{2x} + x - 11 = 0$$ Multiply both sides by x, $$\implies x2^{2x} + x^2 - 11x = 0$$ But with [1], we get $$\frac{11^2}{2^2} + x^2 - 11x = 0$$

I get $$\frac{11 \pm \sqrt{(-11)^2 - 4\frac{11^2}{2^2}}}{2}$$

Which comes to an answer ($$\frac{11}{2}$$) which makes no sense(naturally, since I know I screwed up earlier).

But I did lose track of setting up the $$(a - b)^2 = a^2 - 2ab + b^2$$.

Since $$a^2 = 2^{2x}$$ and $$b^2 = x$$,

Then $$a = 2^{x}$$ and $$b = \sqrt{x}$$.

So $$ab = 2^x\sqrt{x}$$ and $$ab = \frac{11}{2}$$,

Then $$2^x\sqrt{x} = \frac{11}{2}$$

At this point, I'm lost. But just for fun, I took the $$\ln$$ of both sides:

$$x\ln{2} + \frac{1}{2}\ln{x} = \ln{\frac{11}{2}}$$

And finally (without really any reasons to):

$$x + \frac{\ln{x}}{2\ln{2}} = \frac{\ln{\frac{11}{2}}}{\ln{2}}$$

or

$$x + \frac{\ln{x}}{2\ln{2}} = \frac{\ln{11} - \ln{2}}{\ln{2}}$$

And so

$$x + \frac{\ln{x}}{2\ln{2}} = \frac{\ln{11}}{\ln{2}} - 1$$

With no where else to go, I figured I'd ask here. Mind you, I came across a 'similar' question here How to solve the equation:$(2x)^{x}=14+x^{x}$?

It isn't the same, but seeing it uses techniques I don't understand (never learnt the Lambert function), I'm guessing $$4^x + x = 11$$ probably needs the same methods.

• Sadly, if Wolfram is any indication, you're stuck using the Lambert $W$ function, as I discuss here. Commented Jun 7 at 5:32
• @PrincessEev Thanks for the link! While I don't understand the Lambert W function, will try to read up on it (though I suspect it might be over my head). Commented Jun 7 at 6:51

If its using the lambert function, then as per the link given by @PrincessEev, we will use the following property of Lambert $$W$$ function, $$W(x)e^{W(x)}=x=W(xe^x)$$ Now, put $$k=11-x$$, so $$x=11-k$$ and the equation in the problem can be written as, $$e^{(11-k)ln4}=k\implies \frac{e^{11ln4}}{e^{kln4}}=k$$ and so, $$e^{11ln4}ln4=kln4e^{kln4}$$ and applying lambert function on both sides as $$W(xe^x)=x$$, we have, $$k=\frac{W(e^{11ln4}ln4)}{ln4}$$ which can be solved further to get, $$x=11-\frac{W(e^{11ln4}ln4)}{ln4}$$ Proof of the property used:

As, $$W$$ is the inverse of $$f(x)=xe^x$$, its trivial that $$W(xe^x)=x$$ and putting,$$xe^x=k$$, we have, $$W(k)=x \implies W(k)e^{W(k)}=xe^x=k$$ Hence the property is proved.

• Thanks for the answer. But while I have the formula... how does one even get the numerical value of $W(...)$? Commented Jun 7 at 6:53
• Actually I dont know about lambert function myself but used the material provided by Princesseve, and i thought that as he stated there is no other way, so i used it. As far as numerical value of lambert function at a point is concerned i think you can find an approximation using hit and trial only. Commented Jun 7 at 6:56
• Apparently it is needed to use the Lambert Function. Thanks for the help though. My maths isn't up to standard to figure these types of questions. Commented Jun 7 at 6:59
• Thanks for the appreciation but even I never knew of this function and just copied the process which was in the material given by princesseev Commented Jun 7 at 7:02
• Praise the computer. Commented Jun 7 at 9:17

If you need to find the zero of function $$f(x)=4^x + x- 11$$ and that you cannot use Lambert function, only numerical methods would do the job.

But, instead of considering $$f(x)$$, take logarithms and consider $$g(x)=\log(4^x)-\log(11-x)=2x \log(2)-\log(11-x)$$ which is very close to linearity.

By inspection of $$f(x)$$ you know that the solution is somewhere between $$1$$ an $$2$$; so take $$\frac 32$$ as a temporary base point.Using the shortest series expansion around this value $$g(x)=( 4 \log (2)-\log(19))+ \left(\frac{2}{19}+2 \log(2)\right)\left(x-\frac{3}{2}\right)+O\left(\left(x-\frac{3}{2 }\right)^2\right)$$ So, a first estimate is $$x_0=\frac 32+\frac{19 \log \left(\frac{19}{16}\right)}{2 (1+19\log (2))}=1.61522$$ while the exact solution is $$1.61517$$.

Consider that $$x_0 \sim \frac 85$$ and repeat the previous process $$g(x)=\left(\frac{16 \log (2)}{5}+\log (5)-\log(47)\right)+ \left(\frac{5}{47}+2 \log(2)\right)\left(x-\frac{8}{5}\right)+O\left(\left(x-\frac{8}{5}\right)^2\right)$$ then $$x_1=\frac 85-\frac{47 (16 \log (2)+5 \log (5)-5 \log (47))}{5 (5+94 \log (2))}=\color{red}{1.61516}65$$

$$4^x+x=11$$ $$4^x+x-11=0$$

We see, this equation is a polynomial equation of more than one algebraically independent monomials ($$4^x,x$$) and with no univariate factor. We therefore don't know how to solve the equation for $$x$$ by rearranging by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.

To solve the equation in closed form, we therefore have no choice but to use Special functions. As others already wrote, the equation has solutions in terms of Lambert W.