Solve for $x$ in $2\log(x+11)=(\frac{1}{2})^x$

Question

Solve for $x$.

$$2\log(x+11)=(1/2)^x$$

My attempt:

$$\log(x+11)=\dfrac{1}{(2^x)(2)}$$

$$10^{1/(2^x)(2)}= x+11$$

$$x=10^{1/(2^x)(2)}-11$$

I'm not sure what to do next, because i have one $x$ in the exponent while the other on the left side of the equation.

Answer 1

A change of variable can sometimes help clarify what's going on in a formula. In this case it makes sense to try $x+11=u$. The equation becomes

$$\log u={2^{10}\over2^u}$$

With the understanding the "$\log$" means "$\log_{10}$" one can easily observe that both sides equal $1$ when $u=10$, so $x=10-11=-1$ is a solution for the original equation. As Antonio Vargas points out, there are no other solutions: $\log u$ is an increasing function of $u$, while $2^{10}/2^u$ is a decreasing function, so they cross at most once.

It might also help to point out that if that $11$ in the logarithm had been just about any other number, the problem would have been a horrendous mess to solve. In general, the equation $2\log(x+a)=(1/2)^x$ becomes

$$\log u={2^{a-1}\over 2^u}$$

when you let $x+a=u$. If $a=102$ you get $u=100$ as a solution, and if $a=1003$ you get $u=1000$, and so forth, but if $a$ is anything else (integer or rational), then the best you can hope to do is to get approximate values to $u$. There is no nice formula.

Answer 2

Hints:

$$2\log(x+11)=\frac14\implies \log(x+11)=\frac18\implies \color{red}{x+11=e^{1/8}}\ldots$$

Answer 3

For the revised question, assuming $\log = \log_{10}$, the quantity $2 \log(x+11)$ is increasing and the quantity $1/2^x$ is decreasing, so they may only have one point of intersection. Taking $x = -1$ solves the equation, so it is the only solution.

Answer 4

For $x > -11$ assume :

$k1(x) = 2log(x+11)$ which is a continuous, differential, monotonically

increasing function with $k1'(x) > 0$

$k2(x) = (1/2)^x$ which is a continuous, differential, monotonically

decreasing function with $k2'(x) < 0$

$K(x) = k1(x) - k2(x)$ which is a continuous, differential function.

$K'(x) = k1'(x) - k2'(x) > 0$ and so $K(x)$ is a monotonically increasing function.

We solve $K(x) = 0 \iff K(x) = K(-1)$.

K(x) is continuous and monotonic and so the obvious solution $x =-1$ is unique.