Q.290. Logarithm.

Question 290.
If a^(1/3) + b^3(1/3) + c^3(1/3) = 0,
show that log [(a+b+c)/3] = (1/3) (log a + log b + log c).

Answer 290.
a^(1/3) + b^(1/3) + c^(1/3) = 0
=> a^(1/3) + b^(1/3) = - c^(1/3)
 Cubing both sides,
[a^(1/3) + b^(1/3)]^3 = - c
=> [a^(1/3)]^3 + [b^(1/3)]^3 - 3[a^(1/3)] * [b^(1/3)] * [a^(1/3) * b^(1/3)] = - c
=> a + b - 3(ab)^(1/3) * (-c)^(1/3) = - c
=> a + b + c = 3(abc)^(1/3)
=> log [(a+b+c)/3] = (1/3) log(abc)
=> log [(a+b+c)/3] = (1/3) (loga + logb + logc).

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