Question

For any base show that, \[\log \left( 1+2+3 \right)=\log 1+\log 2+\log 3\]. Note that in general, \[\log \left( a+b+c \right)\ne \log a+\log b+\log c\].

Answer 1

Hint: Take LHS and RHS of the given expression separately. Apply basic logarithm formulas and get the expression same in both LHS & RHS.

Complete step-by-step answer:

We have been given the expression, \[\log \left( 1+2+3 \right)=\log 1+\log 2+\log 3\].
Now let us take both LHS and RHS separately.
We know the basic logarithm formula, that is the product rule, \[\log \left( xy \right)=\log x+\log y\].
So we can say that, \[\log 1+\log 2+\log 3=\log \left( 1\times 2\times 3 \right)\].
\[\therefore \log 1+\log 2+\log 3=\log 6-(1)\]
Thus we found the RHS as, \[\log 1+\log 2+\log 3=\log 6\].
Now let us take a look at the LHS of the expression.
\[\log \left( 1+2+3 \right)=\log \left( 3+3 \right)=\log 6\]
Thus we got, \[\log \left( 1+2+3 \right)=\log 6-(2)\].
From (1) and (2) we can say that, \[\log 6=\log 6\], which means that LHS = RHS.
Thus we proved that, \[\log \left( 1+2+3 \right)=\log 1+\log 2+\log 3\].

Note: It is a mere coincidence that 1 + 2 + 3 = 6 and \[\left( 1\times 2\times 3 \right)=6\]. For any other digits , it is not possible to prove like this. You should also know that \[\log \left( a+b+c \right)\] is never equal to \[\log a+\log b+\log c\]. But, \[\log \left( abc \right)=\log a+\log b+\log c\].