Question

Let $f$ be a differentiable function satisfying ${\log }_2(f(3x))=x+{\log }_2(3f(x))\forall x\in R$ and $f^{\prime }(0)=1$
Find the value of $[f(3)]$.
$[$ Note: $[k]$ denotes greatest integer less than or equal to $k$.

Answer 1

Answer: 8

${\log }_2(f(3x))=x+{\log }_2(3f(x))$
$\frac{f(3x)}{3f(x)}=2^x$
$\frac{f(x)}{3\cdot f\left(\frac{x}{3}\right)}\cdot \frac{f\left(\frac{x}{3}\right)}{3f\left(\frac{x}{3^2}\right)}\cdot \frac{f\left(\frac{x}{3^2}\right)}{3\cdot f\left(\frac{x}{3^3}\right)}\dots \dots \cdot \frac{f\left(\frac{x}{3^{n-1}}\right)}{3\cdot f\left(\frac{x}{3^n}\right)}=2^{\frac{x}{3}}\cdot 2^{\frac{x}{3^2}}\dots \dots 2^{\frac{x}{3^n}}$
$\frac{f(x)}{3^n\cdot f\left(\frac{x}{3^n}\right)}=2^{\left(\frac{x}{3}+\frac{x}{3^2}+\dots \dots +\frac{x}{3^n}\right)}$
$\underset{n\to \infty }{\mathrm{Lim}}\left(\frac{f(x)\cdot \frac{x}{3^n}}{f\left(\frac{x}{3^n}\right)\cdot x}\right)=\underset{n\to \infty }{\mathrm{Lim}}{2}^{\left(\frac{x}{3}+\frac{x}{3^2}+\dots \dots +\frac{x}{3^n}\right)}$
$$\begin{gathered} \frac{f(x)}{x\cdot f^{\prime }(0)}=2^{\left(\frac{\frac{x}{3}}{1-\frac{1}{3}}\right)} \\ <br/>f(x)=x\cdot 2^{\left(\frac{x}{2}\right)} \end{gathered}$$
$\therefore f(3)=3\cdot 2^{\left(\frac{3}{2}\right)}=6\sqrt{2}\simeq 8\cdot 4$
$\Rightarrow [f(3)]=8$