smc

While $\log 17=\log (8+9)$, we cannot easily deal with the logarithm of a sum. Furthermore, $17$ is prime, so none of the logarithm rules involving products or differences works. It therefore cannot be found without the use of a table (note: in 1951, calculators were very rare). The correct answer is therefore $\boxed{\log 17}$. As for the rest of the cases: $$\log \frac{5}{4}=\log \frac{10}{8}=\log 10-\log 8=1-\log 8$$ can be found; $$\log 15=\log 3+\log 5=\frac{1}{2}\log 3^2+\log \frac{10}{2}=\frac{1}{2}\log 9+\log 10-\log 2=\frac{1}{2}\log 9+1-\frac{1}{3}\log 8$$ can be found; $$\log 600=\log 100+\log 6=2+\log 2+\log 3=2+\frac{1}{3}\log 8+\frac{1}{2}\log 9$$ can be found; and $$\log .4=\log \frac{4}{10}=\log 4-\log 10=2\log 2-1=\frac{2}{3}\log 8-1$$ can be found.