Ankoganit wrote: Find a real function $f : [0,\infty)\to \mathbb R$ such that $f(2x+1) = 3f(x)+5$, for all $x$ in $[0,\infty)$. For example : $f(x)=-\frac 52$ $\forall x\ge 0$

Does there exist any other function by the way?

Vilakshan wrote: Does there exist any other function by the way? Just set a recurrence ? I mean for $x,y \in [0, \infty)$, define $x \sim y$ if $x = 2^k y + (2^k - 1) y$ for some integer $k$ (or the other way around). This partitions $\mathbb{R}$ into uncountable union of countably infinite sets, and for each set define $f$ for any element arbitrarily and then induct upward and bottomward.

Vilakshan wrote: Does there exist any other function by the way? Oh yes. General solution is $f(x)=3^{\lfloor\log_2(x+1)\rfloor}g(2^{\{\log_2(x+1)\}})-\frac 52$, whatever is $g(x)$ from $[1,2)\to\mathbb R$