Let f : R → R be a function given by f(x + y) = f(x) f(y) for all x, y ∈ R If f(x) ≠ 0 for all x ∈ R and f'(0) = log 2, then f(x) =

$2^x$

We have,

f(x + y) = f(x) f(y) for all x, y ∈ R

⇒ f(0) = f(0) f(0)                       [Replacing x and y both by 0]

⇒ f(0)(f(0) - 1) = 0

⇒ f(0) - 1 = 0

⇒ f(0) = 1

Again,

f(x + y) = f(x) f(y) for all x, y ∈ R

⇒ f(x + h) = f(x) f(h) for all x, h ∈ R

⇒ f(x + h) - f(x) = f(x)(f(h) - 1)

⇒ $\frac{f(x+h)-f(x)}{h}=f(x)\left(\frac{f(h)-1}{h}\right)$

$\Rightarrow \lim\limits_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f(x) \lim\limits_{h \rightarrow 0} \frac{f(h)-f(0)}{h}$                       [∵ f(0) = 1]

⇒ f'(x) = f(x) f'(0) for all x ∈ R

$f'(x)=f(x) . \log _e 2$

⇒ $\frac{f'(x)}{f(x)}=\log _e 2$

$\log _e f(x)=x \log _e 2+\log _e C$            [On integrating]

⇒ $f(x)=C 2^x$

But, f(0) = 1. Therefore, C = 1

Hence, $f(x)=2^x$