$If\;\phi \left( x \right) = \;\phi '\left( x \right)\;and\;\;\phi \;\left( 1 \right) = 2,\;then\;\;\phi \left( 3 \right)\;equals\;$

$\phi(x)=\phi(x)$ and $\phi(1)=2$

Let, $y=\phi(x)$

$\begin{array}{c} y=\frac{d y}{d x} \\ \Rightarrow \quad \int d x=\int \frac{d y}{y} \\ \therefore \quad x=\ln y . \end{array}$

Given,

$\begin{array}{rl} \phi(1) & =2 \\ \Rightarrow 2 & =e^{1+2} \\ \therefore & \ln ^{2}=1+c \\ \Rightarrow c & =\ln ^{2}-1 \\ c & =\ln 2-\ln e[\because \ln e ^{e}=1] \\ c & =\ln 2 / e \\ y & =e^{x+\ln (2 / e)} \\ \phi(3) & =? y=e^{3+\ln (2 / e)}=e^{3} \cdot e^{\ln (2 / e)} \\ & =e^{3} \times \frac{2}{e} \quad\left[\because a^{\log a b}=b\right] \\ & = \phi(3) \\ & =2 e^{2} \end{array}$