please prove 3rd question
Attachments:
Answers
Answered by
0
log(a²/bc)= log a² - log (bc)
= log a + log a - [ log b + log c]
= log a+ log a - log b - log c
similarly we get,
log( b²/ac)= log b+ log b - log a - log c
log(c²/ab) = log c+ log c - log a - log b
therefore
log(a²/bc)+ log(b²/ac)+ log (c²/ab) = log a+log a- log b - log c + log b + log b - log a - log c + log c + log c - log a - log b = 0
HENCE PROVED!
HOPE IT HELPS YOU!
= log a + log a - [ log b + log c]
= log a+ log a - log b - log c
similarly we get,
log( b²/ac)= log b+ log b - log a - log c
log(c²/ab) = log c+ log c - log a - log b
therefore
log(a²/bc)+ log(b²/ac)+ log (c²/ab) = log a+log a- log b - log c + log b + log b - log a - log c + log c + log c - log a - log b = 0
HENCE PROVED!
HOPE IT HELPS YOU!
Similar questions