Problem in exponential/log calculus question

Question

I have no idea how to approach this question, $\frac{dQ}{dt} = Q$ and $Q = e$ when $t = 0$, find $Q$ in terms of $t$. I can approach it logically, and the only way $y' = e$ when $t = 0$ is $y= e^{t+1}$, how do I solve it algebraically?

Answer 1

Assuming that $Q\gt0$ $$\frac{dQ}{dt}=Q$$ By rearranging $$\frac{dQ}{Q}=dt$$ by Integrating both sides we get

Note:$$\int\frac{dx}{x}=\ln|x|+C \text{ and } \int dx =x+C$$

$$\ln Q=t+C$$ $$\ln e=C$$ $$\implies C=1$$ By putting everything back we get

$$\ln Q=t+1 \implies Q=e^{t+1}$$

Answer 2

$\frac{dQ}{dt}=Q \rightarrow \frac{dt}{dQ}=\frac{1}{Q} \rightarrow t=\ln (Q)+c$. Then simply substitute in your initial conditions and you can find Q in terms of t after rearranging.