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
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Answered by
2
concepts : .....(1)
........(2)
.........(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
........(2)
.........(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
Answered by
6
☆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 ☆
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