Question

For any base show that
log (1+2+3) = log 1 + log 2 + log 3.
Note that, in general,
log (a+b+c) ≠ log a + log b + log c

Answer 1

concepts : $log(AB)=logA+logB$.....(1)

$log\frac{A}{B}=logA-logB$........(2)

$logA^n=nlogA$.........(3)


yes of course, in general
log (a+b+c) ≠ log a + log b + log c but here it is possible. let see how it happens.

LHS = log(1 + 2 + 3)

= log(6)

= log(1 × 2 × 3)

= log1 + log2 + log3 = RHS [ using formula (2), ]

hence, it is clear that,
log(1 + 2 + 3) = log1 + log2 + log3

Answer 2

☆your question:-☆

Q) For any base show that log (1+2+3) = log 1 + log 2 + log 3. Note that, in general, log (a+b+c) ≠ log a + log b + log c ?

☆your Answer:-☆

Toprove:-

 log(1+2+3)=log1+log2+log3

Taking

L.H.S.

log(1+2+3)=log6

=log(1×2×3)

=log1+log2+log3

(∵logab=loga+logb)

= R.H.S.

☆Hence proved.☆

☆THANK YOU ☆