why is $\int_0^x f'/f = \log f(x) - \log f(0)$? We have $f' = - cf$, $f$ is defined on the space $[0, \infty)$.
So the solution is somehow:
$$f'/f = -c \Rightarrow \int_0^x f'/f = -cx \Rightarrow \log f(x) = \log f(0) - cx \Rightarrow f(x) = e^{-cx}f(0)$$
So why is $\int_0^x f'/f = \log f(x) - \log f(0)$?
 A: The identity is true only for complex valued functions. For real valued functions the identity have the form
$$\int_0^x \frac{f'(t)}{f(t)}\,\mathrm dt=\ln|f(x)|-\ln|f(0)|$$
The reason is that you can check that the derivative of $\ln |f(x)|$ is $f'(x)/f(x)$ (for real-valued $f$). Then, by the fundamental theorem of calculus, you have the above identity.
A: \begin{align}
\int_0^x \frac{f'(s)}{f(s)} \,ds & = \int \frac{df} f \quad \text{where } df = f'\,ds \\[10pt]
& = \log f.
\end{align}
And as $s$ goes from $0$ to $x$ then $f(s)$ goes from $f(0)$ to $f(s),$ so you get $\log f(s)-\log f(0)$ or $\log\dfrac{f(s)}{f(0)}.$ But the more precise answer is: the chain rule:
$$
\frac d {ds} \log f(s) = \big(\log' f(s)\big) \cdot f'(s) = \frac 1 {f(s)} \cdot f'(s).
$$
A: The only assumption I see in the problem is that $f$ is differentiable on $[0,\infty)$ and $f'=cf$ there. From that, you want to derive $f(x) = f(0)e^{cx}.$ You are trying to reach this conclusion with a dubious argument involving $f'/f$ and $\log f$ and such. It's dubious because $f$ could be $0$ somewhere, so that $f'/f$ is problematic. Also, $f<0$ is a possibility, in which case $\log f$ is problematic.
If we add the assumption that $f>0$ on $[0,\infty)$ at the outset, then $\log f$ is fine there, no problem. We then have $(\log f)' = f'/f$ on $[0,\infty)$ by the chain rule and we have the result by the FTC.
Here's a solution that works in all cases, which I think is simpler: Consider the function $e^{cx}f(x).$ Its derivative is $ce^{cx}f(x) + e^{cx} f'(x) \equiv 0.$ Thus $e^{cx}f(x)$ is constant. The value of the constant is $f(0).$ We conclude $f(x) = f(0)e^{-cx}$ for all $x\in [0,\infty).$
