Prove $\ln{(\frac {x}{y})} = \ln{x} - \ln{y}$ using the definition $\int_1^x\frac{1}{t}dt = \ln{x}$. Prove $\ln (\frac{x}{y}) = \ln{x} - \ln{y}$ using the definition $\int_1^x\frac{1}{t}dt = \ln{x}$.
I am able to prove $\ln{xy} = \ln{x} + \ln{y}$, and $\ln{x^r} = r\ln{x}$, but with this one, I am stuck at $\ln\frac{x}{y} = \ln{x} + \ln{\frac{1}{y}}$, but is it okay to automatically assume $\ln{\frac{1}{y}}$ = $-\ln{y}$ from here?
 A: In case you don't like the $r=-1$ approach, here is a direct proof.
First write $\displaystyle\ln\frac{1}{y} = \int_1^\frac{1}{y}\frac{1}{t}\,\mathrm{d}t = -\int_\frac{1}{y}^1\frac{1}{t}\,\mathrm{d}t$. Now we change variables to $s = yt$. Then $\mathrm{d}s = y\mathrm{d}t$ and thus $\displaystyle\ln\frac{1}{y} = -\int_\frac{1}{y}^1\frac{1}{t}\,\mathrm{d}t = -\int_1^y\frac{1}{s}\,\mathrm{d}s = -\ln y$.
A: You can do the change of variable $u=\frac{1}{t}$: for $x > 0$,
$$
\ln\frac{1}{x} =  \int_{1}^{\frac{1}{x}} \frac{dt}{t} = \int_{1}^{x} \frac{-du}{u^2}u = -\int_{1}^{x} \frac{du}{u} = -\ln x
$$
A: 
I am able to prove $\ln{xy} = \ln{x} + \ln{y}$, and $\ln{x^r} = r\ln{x}$, but with this one, I am stuck at $\ln{x/y} = \ln{x} + \ln{\frac{1}{y}}$, but is it okay to automatically assume $\ln{\frac{1}{y}}$ = $-\ln{y}$ from here?

Yes, absolutely. That's how proofs work in mathematics often work; certain things are proven, and than the results of those earlier proofs can be used in subsequent proofs. 
So do as mathematicians do: just cite the problems in which you established both 


*

*$\,\ln(x\,y) = \ln x + \ln y,\,$ 

*$\,\ln x^r = r \ln x,\,$ 


and you may argue as you suggest.
A: I think you mean $\ln(x/y) = \ln x - \ln y$. If you've already shown that $\ln x^r = r\ln x$ (for all $r\in\Bbb R$), $\ln(1/y) = \ln y^{-1} = -\ln y$ follows from there, so you're good.
