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If $$a = {b^2} = {c^3} = {d^4},$$    then the value of $${\log _a}\left( {abcd} \right)$$   would be -

A. $${\log _a}1 + {\log _a}2 + {\log _a}3 + {\log _a}4$$

B. $${\log _a}24$$

C. $${\text{1 + }}\frac{1}{2} + \frac{1}{3} + \frac{1}{4}$$

D. $${\text{1 + }}\frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{4!}}$$

Answer: Option C

Solution(By Examveda Team)

$$\eqalign{ & a = {b^2} = {c^3} = {d^4} \cr & \Rightarrow b = {a^{\frac{1}{2}}},\,\,\,\,c = {a^{\frac{1}{3}}},\,\,\,\,d = {a^{\frac{1}{4}}} \cr & \therefore {\log _a}\left( {abcd} \right) \cr & = {\log _a}\left( {a \times {a^{\frac{1}{2}}} \times {a^{\frac{1}{3}}} \times {a^{\frac{1}{4}}}} \right) \cr & = {\log _a}{a^{\left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right)}} \cr & = \left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right){\log _a}a \cr & = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \cr} $$

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