Question

If $f\left(x\right)$ is a function such that $f\left(xy\right)=f\left(x\right)+f\left(y\right)$ and $f\left(2\right)=1$ then $f\left(x\right)$

• A. $x^2$
• B. $2^x$
• C. ${\log }_{2}x$
• D. ${\log }_{x}2$

Answer 1

Answer: (C)

${\log }_{2}x$

We have, $f\left(xy\right)=f\left(x\right)+f\left(y\right)\Rightarrow \left(1\right)$
Put $x=y=1$
$\Rightarrow f\left(1\right)=0$
Now $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left[x\right(1+h/x\left)\right])-f(x)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(x\right)+f(1+h/x)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f(1+h/x)-f\left(1\right)}{h/x}\cdot \frac{1}{x}=\frac{f^{\prime }\left(1\right)}{x}$
Now integrating we get, $f\left(x\right)=f^{\prime }\left(1\right)\log x+c$
Since $f\left(1\right)=0\Rightarrow c=0$
Thus $f\left(x\right)=f^{\prime }\left(1\right)\log x$
Also $f\left(2\right)=1\Rightarrow 1=f^{\prime }\left(1\right)\log 2\Rightarrow f^{\prime }\left(1\right)=\frac{1}{\log 2}$
Hence $f\left(x\right)=\frac{\log x}{\log 2}={\log }_{2}x$