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How would I solve $2^x=-2x+11$?

I've tried writing the equation in terms of logarithms, so I did $\log_2 (-2x+11) = x$, but I got stuck.

Then I tried writing it in terms of $be^b$ so I could use the W Lambert function, but that didn't work either. Am I overthinking this? Help!

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    Well Wolfram alpha does give x in terms of the Lambert W so it definitely seems possible – 2017-01-21
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    Actually wikipedia has exactly what you are looking for https://en.m.wikipedia.org/wiki/Lambert_W_function , the examples lead to an equation with a general form of what you have that is $p^x= ax+b$ – 2017-01-21

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As Triatticus commented, this is exactly the first example given in the Wikipedia page. Using the steps, yo should arrive to $$x=\frac{11}{2}-\frac{W\left(16 \sqrt{2} \log (2)\right)}{\log (2)}$$ THe argument being large, you could evaluate $W\left(16 \sqrt{2} \log (2)\right)$ using $$W(a)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\cdots\qquad (L_1=\log(a)\qquad L_2=\log(L_1)$$ given in the same page. Tuncated at this level, you should get $$W\left(16 \sqrt{2} \log (2)\right)\approx 2.04196$$ while the exact value should be $\approx 2.03980$. Applied to $x$, the approximation would give $x\approx 2.55408$ for an excat solution $x\approx 2.55719$. For sure, using more terms would give a better accuracy.

In any manner, if you cannot use Lambert function, solving the equation $$f(x)=x \log(2)-\log(11-2x)=0$$ was a very good idea since it makes the graph almost linear. Using Newton method with $x_0=0$, you would otain the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0 \\ 1 & 2.740560228 \\ 2 & 2.559196364 \\ 3 & 2.557193176 \\ 4 & 2.557192952 \end{array} \right)$$

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    [Winitzki's approximation](http://inis.jinr.ru/sl/_Papers/Computer_algebra/Limits,%20asymptotics/Winitzki.%20Uniform%20approximations%20of%20transcendental%20functions%20(LNCS)(10s).pdf) ought to do pretty well in this case. – 2017-01-21
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    @J.M.isn'tamathematician. For sure, there are many other approximations which works very well. I use many of them based on Pade approximants. – 2017-01-21
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    @LanceLampert. You are very welcome. – 2017-01-21
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Change to natural base.

$$e^{x\ln 2 }=-2x+11$$

Replace $x=-v$

$$e^{-v\ln 2 }=2v+11$$

Now we need to get rid of $+11$ do this by substituting $v=\frac{1}{2}u-\frac{11}{2}$.

$$e^{-\ln 2 (\frac{1}{2}u-\frac{11}{2})}=u$$

$$e^{11\frac{\ln 2 }{2}}e^{-\frac{\ln 2 }{2}u}=u$$

$$e^{11\frac{\ln 2 }{2}}=ue^{\frac{\ln 2 }{2}u}$$

Multiply both sides by $\frac{\ln 2 }{2}$

$$\frac{\ln 2 }{2}e^{\frac{11 \ln 2 }{2}}=\frac{\ln 2 }{2}ue^{\frac{\ln 2 }{2}u}$$

$$W(\frac{\ln 2 }{2}e^{\frac{11 \ln 2 }{2}})=\frac{\ln 2 }{2}u$$

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

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

Back substitute,

$$v=\frac{1}{\ln2}W(\frac{\ln 2 }{2}(\sqrt{2})^{11})-\frac{11}{2}$$

Again,

$$x=\frac{11}{2}-\frac{1}{\ln2}W(\frac{\ln 2 }{2}(\sqrt{2})^{11})$$

Simplifying,

$$x=\frac{11}{2}-\frac{1}{\ln2}W(16 \sqrt{2}\ln 2)$$

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    Wow, this is beautiful! – 2017-01-21