# Variable on both Sides, Exponential = Binomial

How would you solve the following exponential equation:

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

I already tried graphing and "guessing and checking," but I would like to know if it's possible to solve this algebraically.

The answer is approximately $$2.6$$

Thank you for your help.

• Your best bet is a numerical method. Commented Feb 3, 2019 at 2:27
• In general, this kind of problem can't be solved in terms of elementary functions. But you might check out the Lambert-$W$ function. Commented Feb 3, 2019 at 2:44
• Was this a homework problem given in a class, or did you come up with it on your own? If its the first, what class?
– user531621
Commented Feb 3, 2019 at 2:52

## 1 Answer

You ar looking for the zero(s) of function $$f(x)=2^x+2x-11=e^{x \log(2)}+2x-11$$

As said in comments, the solution is given in terms of Lambert function $$x=\frac{11}{2}-\frac{W\left(16 \sqrt{2} \log (2)\right)}{\log (2)}\approx 2.55719$$ The linked page clearly explains the steps and also shows how to evaluate $$W(.)$$ using series.

Notice that $$f'(x)=2^x \log (2)+2$$ never cancels in the real domain making the root to be unique.

I you do not want (or cannot use) this function, the simplest would be Newton method. Plottiing the function, you notice that the solution is close to $$2$$. Then, use $$x_0=2$$ and the iterates would be given by $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.000000000 \\ 1 & 2.628589676 \\ 2 & 2.558377938 \\ 3 & 2.557193278 \\ 4 & 2.557192952 \end{array} \right)$$