Solve for the value of $x$: $2^x=-2x+11 $

Question

Solve for the value of $x$: $$2^x=-2x+11 $$

So this is what I did so far:

I took the log of both sides to obtain:

$$ \log(2^x)=\log(-2x+11)$$ $$ x \log (2)=\log(-2x) \log (11)$$

but now i have a hard time isolating $x$

Any ideas?

Answer 1

One might implement Newton's method to see that

$$a_{n+1}=a_n-\frac{2^{a_n}+2a_n-11}{\ln(2)2^{a_n}+2}$$

with $a_0=2.5$ we find that

$$a_1=2.667953701\\a_2=2.557193086\\a_3=2.557192952\\a_4=2.557192952$$

Thus, the solution is $x\approx2.557192952$

Answer 2

No chance of being able to solve this using elementary functions. You can approximate the solution using a bisection search, or computer algebra tool. Wolfram Alpha reports the answer as approximately $2.55719$.

Answer 3

The solution is given in the Wikipedia of the Lambert W function.

$$x=\frac{11}2-\frac{W\left(\sqrt{2^{9}}\ln(2)\right)}{\ln(2)}$$

Answer 4

So we have,

$$x\ln2=\ln (-2x+11)$$

Or

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

Then consider the recursive equation,

$$x_{n+1}=\frac{\ln (-2x_n+11)}{\ln 2}$$

With $x_1=3$ then if it converges it must converge to one possible value of $x$ because we have $\lim_{n \to \infty} x_n=\lim_{n \to \infty} x_{n+1} :=x$ if it converges. One can show this is the only real value of $x$.

Here are some iterations:

$$x_2=2.32192809....$$

$$x_3=2.66815176....$$

$$x_4=2.50174394...$$

$$x_5=2.58412359...$$

$$x_6=2.5439295...$$

$$x_7=2.56368064...$$

..

$$x_{32}=2.557192952...$$

Going back to your equation,

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

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

Let $-u=x \ln 2 $

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

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

Not let $u=t+b$.

$$1=(\frac{2}{\ln 2}t+\frac{2}{\ln 2}b+11)e^{t+b}$$

And let,

$$\frac{2}{\ln 2}b+11=0 \implies b=-\frac{11\ln 2}{2}$$

Then we have,

$$1=\frac{2}{\ln 2}te^{t-\frac{11\ln 2}{2}}$$

$$1=\frac{2}{\ln 2}e^{-\frac{11\ln 2}{2}}te^t$$

$$\frac{\ln 2}{2}e^{\frac{11\ln 2}{2}}=te^t$$

$$\frac{\ln 2}{2}(e^{\ln 2})^{\frac{11}{2}}=te^t$$

$$\frac{\ln 2}{2} 2^{\frac{11}{2}}=te^t$$

$$2^{\frac{9}{2}}\ln 2=te^t$$

Now we use the lambert W function.

$$W(2^{\frac{9}{2}}\ln 2)=t=u+\frac{11\ln 2}{2}$$

$$W(2^{\frac{9}{2}}\ln 2)-\frac{11\ln2}{2}=u=-x\ln 2$$

$$\frac{11\ln 2}{2}-W(2^{\frac{9}{2}}\ln 2)=x\ln 2$$

$$x=\frac{11}{2}-\frac{W(2^{\frac{9}{2}}\ln 2)}{\ln 2}$$

Answer 5

The derivative of $f(x):=2^x+2x-11$ is $2^x\ln2+2$, which doesn't cancel. Thus the function is monotonic and has at most one root.

By trial and error, $f(2)<0$ and $f(3)>0$ so that there is indeed a root, located in $(2,3)$.

You can estimate the root by numerical methods and Newton should have the fastest convergence rate. (A closed-form expression with the Lambert function is also possible, but then you need to be able to evaluate it.)

We can obtain a first good approximation by using the Taylor development to the second degree and solving the quadratic equation.

$$f(x)\approx4(1+\ln2(x-2)+\frac{\ln^22}2(x-2)^2)+2x-11=0,$$

giving the root $x=2.56444\cdots$.