Value of 1/(log_aabc)+1/(log_babc)+1/(log_cabc)I guess the answer will be 1. But I don't know how to evaluate it. Can someone give me some tips?

MMDCCC50m

MMDCCC50m

Answered question

2022-11-13

Value of 1 log a a b c + 1 log b a b c + 1 log c a b c
I guess the answer will be 1. But I don't know how to evaluate it. Can someone give me some tips?

Answer & Explanation

Savanna Smith

Savanna Smith

Beginner2022-11-14Added 17 answers

Hint:
log x y = ln y ln x
Thus
log a a b c = ln a b c ln a = ln a + ln b + ln c ln a
pin1ta4r3k7b

pin1ta4r3k7b

Beginner2022-11-15Added 3 answers

I don't know whether am i correct or not.
1 log a a b c + 1 log b a b c + 1 log c a b c = 1 ln a b c ln a + 1 ln a b c ln b + 1 ln a b c ln c
= ln a ln a b c + ln b ln a b c + ln c ln a b c
= ln a + ln b + ln c ln a b c = ln ( a b c ) ln a b c = 1

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