Condition to bound the decay rate of an exponentially stable linear system using Lyapunov stability theory? Could someone elaborate or provide additional justification for a bound on the decay rate of an exponentially stable system?  The following comes from Boyd, El Ghaoui, Feron, and Balakrishan's book, "Linear Matrix Inequalities in System and Control theory" on page 66:

For reference, the definition of LDI (5.1) is found on page 61:

If $dV(x)/dt \leq -2\alpha V(x)$, then why does $V(x(t)) \leq V(x(0))e^{-2\alpha t}$?  I know that from basic Lyapunov theory, $V(x(t))=V(x(0)) + \int_0^t \dot{V}(x(\tau))d\tau\leq V(x(0))$ since $\dot{V}(x(t))\leq 0$.  The above has the extra exponential that I don't know how to derive.  Second, if $V(x(t)) \leq V(x(0))e^{-2\alpha t}$, then why does $\|x(t)\|\leq e^{-\alpha t}\kappa(P)^{1/2}\|x(0)\|$?
 A: @mastrok helped bring me to the final solution.
Let $J(t)=\log(V(x(t))$.  Then, we have that
$$
J^\prime(t) = \frac{1}{V(x(t))} V^\prime(x(t)) x^\prime(t)\leq \frac{1}{V(x(t)} (-2\alpha V(x(t))) = -2\alpha
$$
Next, we have that
$$\begin{align}
\log(V(x(t)))&=\log(V(x(0)))+\int_0^t J^\prime(t) d\tau\\
&\leq\log(V(x(0)))-\int_0^t 2\alpha\\
&=\log(V(x(0)))- 2\alpha t\\
&= \log(V(x(0))e^{-2\alpha t})
\end{align}$$
This implies $V(x(t))\leq V(x(0))e^{-2\alpha t}$ since $\log$ is monotonically increasing and this answers part one of my question above.
For part 2, note that
$$\begin{align}
&&\lambda_{\min}(P)\|x(t)\|^2
&\leq x(t)^T P x(t)\\
&&&\leq x(0)^T P x(0) e^{-2\alpha t}\\
&&&\leq \|x(0)\|^2 \lambda_{\max}(P) e^{-2\alpha t}\\
\Longrightarrow
&&\|x(t)\|^2
&\leq \frac{\lambda_{\max}(P)}{\lambda_{\min}(P)}\|x(0)\|^2e^{-2\alpha t}\\
\Longrightarrow
&&\|x(t)\|
&\leq \sqrt{\frac{\lambda_{\max}(P)}{\lambda_{\min}(P)}}\|x(0)\|e^{-\alpha t}\\
&&&=\kappa(P)^{1/2}\|x(0)\|e^{-\alpha t}
\end{align}$$
