CatalinBordea wrote: Solve in the set of real numbers the equation $ a^{[ x ]} +\log_a\{ x \} =x , $ where $ a $ is a real number from the interval $ (0,1). $ $ [] $ and $ \{\} $ denote the floor, respectively, the fractional part. Setting $x=n+y$ where $n\in\mathbb Z$ and $y\in(0,1)$, equation is $a^n+\frac{\ln y}{\ln a}=n+y$ And so $y-\frac{ln y}{\ln a}=a^n-n$ Which always have one real root $y\in(0,1)$ whatever is $n>0$ (and no root in $'0,1)$ when $n\le 0$) Setting $y=e^t$ with $t<0$ and $a^n-n=-u$ with $u>0$, this is $e^t-\frac t{\ln a}=u$ Whose solution is easily got : $t=-u\ln a-W(-a^{-u}\ln a)$ And so $\boxed{x=n+a^{n-a^n}e^{-W(-a^{n-a^n}\ln a)}}$ whatever is positive integer $n$ (where $W(x)$ is the Lambert W function)

With some late attention, itr seems that solution can be simplified as $\boxed{x=n+a^n}$ whatever is positive integer $n$