I have found this weird equation in a math book. Could you give me any hints? $${x}^{\lfloor x \rfloor} = a$$ for a given a. I have dealt with the trivial cases where $x$ is an integer but cannot solve it in $\mathbb{R}$.
Graph of $x\mapsto {x}^{\lfloor x \rfloor}$ 
I'll show how to find solutions $x>0$.
Note that for $0<x<1$ we have $\lfloor x\rfloor = 0$ so $f(x) = x^0 = 1$, and for $x\ge1$ the funcion $f(x) =x^{\lfloor x\rfloor}$ is increasing - that is because if $1\le x<y$ then $1<\lfloor x\rfloor \le \lfloor y\rfloor$ and $$ x^{\lfloor x\rfloor} < y^{\lfloor x\rfloor} \le y^{\lfloor y\rfloor}$$ The function is also piecewise continuous; specifically, it is continuous on every interval $[n,n+1)$, $n\in\mathbb{N}$ and has discontinuiities at $n\in\mathbb{N}$, $n\ge 2$ that is because $$\lim_{x\to n_-} f(x) = \lim_{x\to n_-} x^{n-1} = n^{n-1} $$ $$\lim_{x\to n_+} f(x) = \lim_{x\to n_+} x^{n} = n^{n} $$ If $n > 1$ it means that there's a discontinuity at $n$. We can therefore conclude that
If $a = 1$ then equation $f(x) = a$ is satisfied by all $x\in(0,1]$.
If $a>1$ and $a \in [n^n,(n+1)^n)$ for some $n\in\mathbb N$ then equation $f(x) = a$ has one solution $x>0$, and this solution satisfies $x\in[n,n+1)$ , therefore we have $\lfloor x \rfloor =n$ and we have $x^n=a$, that is $x=\sqrt[n]{a}$.
If $a<1$ or $a \in [(n+1)^n,(n+1)^{n+1})$ for some $n\in\mathbb N$ then equation $f(x) = a$ has no solutions $x>0$.
I'll leave the searching for solutions $x<0$ to you, it can be done in a similar fashion.