Exponential Function with Variable on Both Sides 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!
 A: 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
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 0 \\
 1 & 2.740560228 \\
 2 & 2.559196364 \\
 3 & 2.557193176 \\
 4 & 2.557192952 
\end{array}
\right)$$
A: 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,
$$x=\frac{11}{2}-\frac{1}{\ln2}W(16 \sqrt{2}\ln 2)$$
