Prove that 1/logab abc + 1/logbc abc +1/logca abc=2.

Question

Prove that \(\cfrac{1}{log_{ab}\,abc}+\cfrac{1}{log_{bc}\,abc}+\cfrac{1}{log_{ca}\,abc}=2.\)

Answer 1

L.H.S. = \(\cfrac{1}{log_{ab}\,abc}+\cfrac{1}{log_{bc}\,abc}+\cfrac{1}{log_{ca}\,abc}\)

= log_abc ab + log_abc bc + log_abc ca

= log_abc (ab\(\times\)bc\(\times\)ca)

= log_abc (a²b²c²)

= log_abc (abc)²

= 2 log_abc abc

= 2 \(\times\) 1 ….[\( \because\) log_a a = 1]

= 2

= R.H.S.