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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.

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  • $\begingroup$ Your best bet is a numerical method. $\endgroup$
    – Sean Roberson
    Feb 3, 2019 at 2:27
  • $\begingroup$ In general, this kind of problem can't be solved in terms of elementary functions. But you might check out the Lambert-$W$ function. $\endgroup$
    – Gerry Myerson
    Feb 3, 2019 at 2:44
  • $\begingroup$ 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? $\endgroup$
    – user531621
    Feb 3, 2019 at 2:52

1 Answer 1

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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 $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.000000000 \\ 1 & 2.628589676 \\ 2 & 2.558377938 \\ 3 & 2.557193278 \\ 4 & 2.557192952 \end{array} \right)$$

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