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.
$\endgroup$ 54 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$Hints:
$$2\log(x+11)=\frac14\implies \log(x+11)=\frac18\implies \color{red}{x+11=e^{1/8}}\ldots$$
$\endgroup$ 3 $\begingroup$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.
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