Taking the logarithmic derivative of an exponential difference function after applying L'Hospital's RuleCan somebody please...

Taking the logarithmic derivative of an exponential difference function after applying L'Hospital's Rule
Can somebody please explain the following application of L’Hospital’s Rule?
Find the limit:
$\underset{x\to 0}{lim}\frac{5^x-3^x}{x}$
Solution:
Determining that this function has indeterminate form $0/0$, we apply L’Hospital’s rule.
Applying L.H. rule, we get $\underset{x\to 0}{lim}\frac{5^x\ln 5-3^x\ln 3}{1}=\underset{x\to 0}{lim}(\ln 5-\ln 3)=\ln \frac{5}{3}$
The part that I am confused on is the application of the natural logarithm that occurs after the first application of L’Hospital’s Rule, namely $\underset{x\to 0}{lim}\frac{5^x\ln 5-3^x\ln 3}{1}=\underset{x\to 0}{lim}(\ln 5-\ln 3)$. I am not sure how the derivative becomes what it does, nor do I understand the next step. I do understand $\underset{x\to 0}{lim}(\ln 5-\ln 3)=\ln \frac{5}{3}$, however.
Thanks!