A property of logarithms is that log(x*y) = log(x) + log(y)
Another is that log(x-1) = -log(x)
So: log(2*a*b) = log(2) + log(a) + log(b)
log(5/a) = log(5) + log(1/a) = log(5) - log(a) (as 1/a = a-1)
log(10*b) = log(10) + log(b)
Therefore: log(2*a*b) + log(5/a) - log(10*b) = log(2) + log(a) + log(b) + log(5) - log(a) - log(10) - log(b)
= log(2) + log(5) - log(10)
= log(10) - log(10)
= 0
I see Melody beat me to it (with a slightly different approach)!