Question

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

Answer 1

$\log _2(f(3 x))=x+\log _2(3 f(x))$
$\frac{f(3 x)}{3 f(x)}=2^x$
$\frac{f(x)}{3 \cdot f\left(\frac{x}{3}\right)} \cdot \frac{f\left(\frac{x}{3}\right)}{3 f\left(\frac{x}{3^2}\right)} \cdot \frac{f\left(\frac{x}{3^2}\right)}{3 \cdot f\left(\frac{x}{3^3}\right)} \ldots \ldots \cdot \frac{f\left(\frac{x}{3^{n-1}}\right)}{3 \cdot f\left(\frac{x}{3^n}\right)}=2^{\frac{x}{3}} \cdot 2^{\frac{x}{3^2}} \ldots \ldots 2^{\frac{x}{3^n}}$
$\frac{f(x)}{3^n \cdot f\left(\frac{x}{3^n}\right)}=2^{\left(\frac{x}{3}+\frac{x}{3^2}+\ldots \ldots+\frac{x}{3^n}\right)}$
$\underset {n \rightarrow \infty}{\text{Lim}}\left(\frac{f(x) \cdot \frac{x}{3^n}}{f\left(\frac{x}{3^n}\right) \cdot x}\right)=\underset {n \rightarrow \infty}{\text{Lim}} 2^{\left(\frac{x}{3}+\frac{x}{3^2}+\ldots \ldots+\frac{x}{3^n}\right)}$
$\frac{f(x)}{x \cdot f^{\prime}(0)}=2^{\left(\frac{\frac{x}{3}}{1-\frac{1}{3}}\right)} \\
f(x)=x \cdot 2^{\left(\frac{x}{2}\right)} $
$\therefore f(3)=3 \cdot 2^{\left(\frac{3}{2}\right)}=6 \sqrt{2} \simeq 8 \cdot 4 $
$\Rightarrow[f(3)]=8$