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Deduce them (that is,  deduce: ln(a * b) = ln(a) + ln(b) , and (log_ab ) *(log_bc) = log_ac.

Solution:

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(a * b))  on the left, which is a * b, and exp(ln(a) + ln(b)) on the right.

Using  the identity  exp(s + t) = exp(s) * exp(t)  the right hand side becomes a * b as well.

The identity exp(s * t) = (exp(s))^t implies the second identity in the case a = exp(1) by the following argument:
in that case log_ab is ln b and log_ac = 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) * log_bc) = exp(ln c) = c

By the identity the left hand side here becomes exp(ln(b))^log_bc, or b ^log_bc which is again c.

But this tells us that in general

and the general claim, log_ab * log_bc = log_ac follows immediately from this fact, by substitution.