Solve this

Given that $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ such that $\mathrm{f}(\mathrm{x})=2^{\mathrm{x}}$

To find: (i) Range of $x$

Here, $f(x)=2^{x}$ is a positive real number for every $x \in R$ because $2^{x}$ is positive for every $x$ ER.

Moreover, for every positive real number $x, \exists \log 2^{x} \in R$ such that

$f\left(\log _{2} x\right)=2^{\log _{2} x}$

$=x\left[\because a^{\log _{a} x}=x\right]$

Hence, the range of f is the set of all positive real numbers.

To find: (ii) $\{x: f(x)=1\}$

We have, $f(x)=1 \ldots(a)$

and $f(x)=2^{x} \ldots(b)$

From eq. (a) and (b), we get

$2^{x}=1$

$\Rightarrow 2^{x}=2^{0}\left[\because 2^{0}=1\right]$

Comparing the powers of 2, we get

$\Rightarrow x=0$

$\therefore\{x: f(x)=1\}=\{0\}$

Comparing the powers of 2, we get

$\Rightarrow x=0$

$\therefore\{x: f(x)=1\}=\{0\}$

To find: (iii) $f(x+y)=f(x) . f(y)$ for all $x, y \in R$

We have,

$f(x+y)=2^{x+y}$

$=2^{x} \cdot 2^{y}$

[The exponent "product rule" tells us that, when multiplying two powers that have the same base, you can add the exponents or vice - versa]

$=f(x) \cdot f(y)\left[\because f(x)=2^{x}\right]$

$\therefore f(x+y)=f(x) . f(y)$ holds for all $x, y \in R$