log(a3b3/c3) + log(b3c3/d3) + log(c3d3/a3) - 3logb2c
Solution
Correct Answer: Option A
log(a3b3/c3) + log(b3c3/d3) + log(c3d3/a3) - 3logb2c
= log[(a3b3/c3) × (b3c3/d3) × (c3d3/a3)] - 3logb2c
= logb6c3 - 3logb2c
= log(b2c)3 - 3logb2c
= 3logb2c - 3logb2c
= 0
= log[(a3b3/c3) × (b3c3/d3) × (c3d3/a3)] - 3logb2c
= logb6c3 - 3logb2c
= log(b2c)3 - 3logb2c
= 3logb2c - 3logb2c
= 0
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