I know this question isn't very difficult but I'm not convinced I'm doing it right.
For a discrete random variable if you have the CDF, the pdf is defined as $f(x)=F(x)-F(x-)$.
I have:
$$F(x) =\begin{cases}0,&{x\le 0}\\log(x+1),&{x=1,2,3,...,9}\\1,&{x\gt 9}\\\end{cases}$$
What I did was:
$$f(x)=F(x)-F(x-)$$ $$=log(x+1)-log(x)$$ $$=log\left(\frac{x+1}{x}\right)$$
Making the pdf: $f(x) =log\left(\frac{x+1}{x}\right)$ for $x=1,2,3,...,9$
This was so trivial I'm in the mind set that I must be missing some detail. Is this correct?
Your solution is correct. You can check this by the telescoping sum: $\sum_{k=1}^{9} \log_{10} (\frac{k+1}{k}) = \log_{10}10 = 1$. Hence this is a probability distribution and each $0<p_k<1$