Proof of natural log identities I need to prove a few of the following identities from a real analysis perspective- this means I do not have access the $\ln e^2 = 2$ type definition of the log function. I am developing the log function from the definition $log x = \int_1^x \frac1t \mathrm dt$ for $0 < x$.
I need to prove that: $\ln(ab) = \ln(a) + \ln(b)$ first. I imagine the proof for that will apply itself pretty directly to $\ln(x^n) = n\ln(x)$. I've been given the hint that the idea here is to define $f(x) = \ln(ax)$ and show that $f'(x) = \ln(x)'$, implying $f(x) = L(ax) = L(x) + k$, where I can show that $k = L(a)$. 
Any help with starting this?
 A: Let $f(x) = \ln(x)$ and let $g(x) = \ln(ax)$, where $a$ is some constant. We claim that $f$ and $g$ only differ by a constant. To see this, it suffices to prove that they have the same derivative. Indeed, by the fundamental theorem of calculus and chain rule, we have that:
\begin{align*}
g'(x) &= \frac{d}{dx} \ln(ax) 
= \frac{d}{dx} \int_1^{ax} \frac{dt}{t} 
= \frac{1}{ax} \cdot a 
= \frac{1}{x} 
= \frac{d}{dx} \int_1^{x} \frac{dt}{t} 
= \frac{d}{dx} \ln(x) 
= f'(x)
\end{align*}
Hence, we have that $g(x) = f(x) + k$ for some constant $k$. Now substitute $x=1$. This yields:
\begin{align*}
g(1) &= f(1) + k \\
\ln(a \cdot 1) &= \ln(1) + k \\
\int_1^{a \cdot 1}\frac{dt}{t} &= \int_1^1\frac{dt}{t} + k \\
\int_1^{a}\frac{dt}{t} &= 0 + k \\
\ln(a) &= k \\
\end{align*}
So we have that $\ln(ax) = \ln(a) + \ln(x)$, as desired.
A: Shorter version
$$\ln(ab)=\int_1^{ab}\frac{1}{t}dt=\int_b^{ab}\frac{1}{t}dt+\int_1^{b}\frac{1}{t}dt=\int_1^{a}\frac{1}{u}du+\int_1^{b}\frac{1}{t}dt=\ln(a)+\ln(b)$$
where I made the substitution $u=bt$.
