$\text { If } \dfrac{d y}{d x}=y+3>0 \text { and } y(0)=2 \text { then } y(\log 2) \text { is equal to }$

The given differential equation is $\dfrac{dy}{dx} = y + 3$.

Rearranging the terms to separate variables, we get $\dfrac{dy}{y + 3} = dx$.

Integrating both sides, $\int \dfrac{dy}{y + 3} = \int dx$, which gives $\ln|y + 3| = x + C$.

Given $y(0) = 2$, substituting $x = 0$ and $y = 2$ into the equation:

$\ln|2 + 3| = 0 + C \Rightarrow C = \ln 5$.

Thus, $\ln(y + 3) = x + \ln 5$.

Exponentiating both sides, $y + 3 = 5e^x$, so $y = 5e^x - 3$.

To find $y(\ln 2)$, substitute $x = \ln 2$:

$y(\ln 2) = 5e^{\ln 2} - 3 = 5(2) - 3 = 10 - 3 = 7$.

Answer: 7