Solve for $k$ the following equation I need to find the value of $k$ for which the following holds.
$$2^{\frac k2 + 1} = 3^k + 1$$
and $k\in\mathbb{R}$.
But I don't even know where to start.
 A: For $k=0$ the equation holds.
Now define
$$f(x)=3^x-2^{\frac x2+1}+1$$
Then
$$f'(x)=3^x\ln 3-\frac12 2^{\frac x2+1}\ln 2=3^x\ln 3- (\sqrt 2)^x\ln 2$$
But
$$\frac{3^x\ln3}{(\sqrt 2)^x\ln 2}=\left(\frac 3{\sqrt 2}\right)^x\log_23$$
So $f'$ is positive for some interval $(c,\infty)$ and negative for $(-\infty,c)$.
We see also that $c<0$ because $f'(0)=\ln 3-\ln 2>0$.
Namely, 
$$c=\log_{3/\sqrt 2}\log_32\approx -0.612$$
To see if there is a solution in $(-\infty,c)$, we compute
$$\lim_{x\to-\infty}f(x)=1-1+1=1$$
so we see that $f(x)>0$ at some interval $(-\infty, k)$ and $f(-1)=\frac43-\sqrt 2<0$, so, by Bolzano's theorem, there is another solution.
There are no more solutions because $f'$ vanishes only at one point.
A: Since solving does not seem to be easy, you might look what happens for simple values. It turns out that $$2^{\frac02+1}=2^1=3^0+1=1+1.$$
The graph below, however, suggests there is another solution. It's approximately $-1.484$ but I don't really see how to get there manually either. Plus, I have this Symbolab application in which I tried to plug in this equation, but even with that I don't get a nice solution.
If you're asked for the value of $k$, I think the question is just wrong since there are two solutions.

A: There are two (real) solutions: $x=0$ and $x\approx -1.48378$.
This can be seen by studying the function $f(x) = 2^{x/2+1}- 3^x-1$. Its derivative is $f'(x) = 2^{x/2} \log(2) - 3^x \log(3)$ which has one root
$$x_0= -2\frac{ \log(\log 2 ) - \log(\log 3)}{\log(2) - 2 \log 3} \approx -0.6124.$$
By checking at some value on each side we notice that $f'(x) > 0$, when $x<x_0$ and $f'(x)<0$ when $x>x_0$ (by continuity of $f'$).
Hence $f$ has a local maximum at $x_0$ and is strictly monotonic on each side. By checking, we see that $f(x_0)>0$ and also that $f(x_1)<0$ for some $x_1 < x_0$ and similarly that $f(x_2)<0$ for some $x_2 > x_0$ (take for example $x_1=-2$ and $x_2 = 1$). By continuity, $f$ has exactly $2$ roots.
As is easily notices, the second solution (on the right side of $x_0$) is $0$. The first solution can be approximated, but I don't know if one can get a nice algebraic expression for it.
