Math, asked by gonilan6015, 1 year ago

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

Answers

Answered by abhi178
2
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
Answered by ravindrabansod26
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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