What is $\log(n+1)-\log(n)$?

$\log(n+1)-\log n=\log(1+\frac1n)$. Using the Taylor series for $\log(1+x)$, this is $$\frac1n-\frac1{2n^2}+\frac1{3n^3}-\cdots\approx\frac1n.$$

Especially Lime's answer is absolutely correct and a very good approach. Let me show a conceptually somewhat simpler one, that only uses the derivative, namely $(\log x)'=1/x$. This is a strictly decreasing function, so $\log x$ is concave. In particular, its graph is below the tangent line at any point of the graph.

We obtain $\log (n+1)< \log n + 1/n$ and $\log n< \log (n+1) - 1/(n+1)$. To sum up:

$$\frac{1}{n+1} < \log (n+1)-\log n < \frac{1}{n}$$

This is rather similar to previous answers, but I think it's still worth pointing out. You're asking about the slope of a chord of the graph of $\log x$, the chord joining $(n,\log n)$ to $(n+1,\log(n+1))$. By the mean value theorem, this equals the slope of the tangent line, $1/x$, at some $x$ between $n$ and $n+1$.