Deduce them, that is, deduce: ln ab = ln a + ln b, and (logab ) *(logbc) = logac.
The natural logarithm function ln x is inverse to the exponential function. If in the first equation we take the exponential function of both sides we get exp(ln ab) on the left, which is ab, and exp(ln a + ln b) on the right.
Using the identity exp(s + t) = (exp s) * (exp t) the right hand side becomes ab as well.
The identity exp st = (exp s)t implies the second identity
in the case a = exp 1 by the following argument:
in that case logab is ln b and logac = ln c.
We can then apply the exponential function to both sides of the equation we want to prove and we get
exp((ln b) * (logbc)) = exp(ln c) = c
By the identity the left hand side here becomes exp(ln b)^logbc, or b ^logbc which is again c.
But this tells us that in general
and the general claim, (logab) * (logbc) = logac follows immediately from this fact, by substitution.