Is there an algebraic solution to $\log_{\sqrt2}{\left(x\right)} = (\sqrt2)^x$? I’m trying to solve
$$\log_{\sqrt2}{\left(x\right)} = (\sqrt2)^x$$
My next step is
$$\ln{x}= (\sqrt2)^x\ \cdot\ \ln\sqrt2$$
EDIT:
I’m only up to high school math.
 A: Note that $(\sqrt2)^x=(e^{\ln\sqrt2})^x = e^{x\ln\sqrt2}$, and $\ln\sqrt2=\ln{(2^{1/2})}=\frac12\ln2$
That said, it may be easier to go back a step and use 2 as our base instead of $e$:
Starting with $\log_{\sqrt{2}}(x)=(\sqrt{2})^{x}$, on the left-hand side we have
$$\log_{\sqrt{2}}(x)=\frac{\log_{2}(x)}{\log_{2}(\sqrt2)}=\frac{\log_{2}(x)}{1/2}= 2 \log_{2}(x)=\log_{2}(x^2)$$
(I've introduced an extraneous solution in that last step, but if we assume $x>0$ we'll be fine.)
And on the right-hand side we have $(\sqrt{2})^{x}=(2^{1/2})^x=2^{\ x/2}$, so
$$\log_{2}(x^2)=2^{\ x/2}\Rightarrow x^2 = 2^{2^{(x/2)}}\Rightarrow x = \sqrt{2^{2^{(x/2)}}}=(2^{2^{(x/2)}})^{1/2}$$
... and at this point I'm stumped. I know the answer has to do with the fact that $2^2=2\cdot 2=2+2=4$, but my closest attempt at getting WolframAlpha to solve it by algebraic manipulations has resulted in expressions which involve the Lambert W function but somehow evaluate to 2 and 4.
Maybe it might be useful to note that to note is that $f(x)=\log_{\sqrt{2}}(x)$  and $g(x)=(\sqrt{2})^{x}$ are inverse functions of each other, so when $f(x)=g(x)$, then they are both equal to $x$. From there the path seems to be through the Lambert W function, which cannot be expressed in terms of elementary functions.
I don't feel like this is an answer yet, but I want to post my work so far in case it helps.
A: $$\log_{\sqrt{2}}(x)=\left(\sqrt{2}\right)^x$$
$$\frac{\ln(x)}{\ln(\sqrt{2})}=\left(\sqrt{2}\right)^x$$
$$\frac{\ln(x)}{\ln(\sqrt{2})}-\left(\sqrt{2}\right)^x=0\tag{1}$$
$F(x)=\frac{\ln(x)}{\ln(\sqrt{2})}-\left(\sqrt{2}\right)^x$:
$$F(x)=0$$
We see from (1), your equation is an algebraic equation in dependence of two algebraically independent monomials ($\frac{\ln(x)}{\ln(\sqrt{2})}$, $\left(\sqrt{2}\right)^x$). With help of the main theorem in [Ritt 1925], that is also proved in [Risch 1979], we can conclude that  the elementary function $F$ doesn't have a partial inverse that is an elementary function. Therefore, the equation cannot be rearranged for $x$ by applying only elementary functions/operations we can read from the equation.
[Risch 1979] Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759
[Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90
a)
Because the real functions $f$ and $g$ are the inverses of each other, as ReGuess writes in his answer, the problem in the reals can be splitted into a simpler problem: it's enough to solve one of the following two equations. The graphs of $f$ and $g$ together with the graph of the identity function show this for the reals.
$$\frac{\ln(x)}{\ln(\sqrt{2})}=x$$
$$\left(\sqrt{2}\right)^x=x$$
Each of these equations can be solved by Lambert W, as ReGuess correctly writes. Each of the equations has the real solutions $2$ and $4$.
b)
an extra for the interested reader:
Your equation can be solved in terms of generalized hyper-Lambert (see the references below):
$$\frac{2\ln(x)}{\ln(2)}=e^{\frac{1}{2}\ln(2)x}$$
$$\frac{2\ln(x)e^{-\frac{1}{2}ln(2)x}}{\ln(2)}=1$$
$$\ln(x)e^{-\frac{1}{2}\ln(2)x}=\frac{1}{2}\ln(2)$$
$x=e^t$:
$\forall k_1\in\mathbb{Z}$:
$$te^{-\frac{1}{2}\ln(2)e^t}=\frac{1}{2}\ln(2)+2k_1\pi i$$
$$G(-\frac{1}{2}\ln(2);t)=\frac{1}{2}\ln(2)+2k_1\pi i$$
$$t=HW(-\frac{1}{2}\ln(2);\frac{1}{2}\ln(2)+2k_1\pi i)$$
$\forall k_2\in\mathbb{Z}$:
$$x=\ln(HW(-\frac{1}{2}\ln(2);\frac{1}{2}\ln(2)+2k_1\pi i))+2k_2\pi i$$
Galidakis, I. N.: On solving the p-th complex auxiliary equation $f^{(p)}(z)=z$. Complex Variables 50 (2005) (13) 977-997
Galidakis, I. N.: On some applications of the generalized hyper-Lambert functions. Complex Variables and Elliptic Equations 52 (2007) (12) 1101-1119
A: $\log_{\sqrt{2}}(x)$ is concave down. $\sqrt{2}^x$ is concave up. Therefore there are at most two solutions.
After writing these more "nicely" as $2\log_2(x)$ and $2^{x/2}$, we can "see" that $2$ and $4$ are solutions, by luck of these particular numbers.
A: The equation as presented takes the form
$$ 2 \, \ln x = 2^{x/2} \, \ln 2. $$
It is fairly clear that if $x = 2$ then $ 2 \, \ln 2 = 2 \, \ln 2$ which is a solution to the equation. The next solution should be an even integer value, like $x = 4$.
A: This may be of interest:
$$\log_{\sqrt 2}(x) = (\sqrt{2})^x$$
$$\implies {\sqrt{2}}^{\log_{\sqrt 2}(x)} = {\sqrt{2}}^{(\sqrt{2})^x}$$
$$\implies x = {\sqrt{2}}^{(\sqrt{2})^x}$$
$$\implies  x = 2^{\frac{1}{2} \times 2^{\frac{x}{2}}}.$$
The LHS is linear, whereas the RHS is doubly exponential and convex [and rapidly increasing in $x$]. There is no nice way in general to solve this, but by inspection $x=2$ and $x=4$ works [as already noted in the previous answer].
