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If $a=b^{2}=c^{3}=d^{4}$ then the value of $\log_{a}\;(abcd)$ would be

  1. $\log_{a}1+\log_{a}2+\log_{a}3+\log_{a}4$
  2. $\log_{a}24$
  3. $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$
  4. $1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}$
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Given that $:a = b^{2} = c^{3} = d^{4}$

we can take $a = b^{2}\implies \log_{a}b = \dfrac{1}{2}$

Similarly, $a = c^{3}\implies \log_{a}c = \dfrac{1}{3}$

and , $a = d^{4}\implies \log_{a}d = \dfrac{1}{4}$

Now, $\log_{a}(abcd) = \log_{a}a +\log_{a}b +\log_{a}c +\log_{a}d $

$\implies \log_{a}(abcd) = 1 + \dfrac{1}{2}+ \dfrac{1}{3}+ \dfrac{1}{4}$

Reference:https://brilliant.org/wiki/logarithms/

So, the correct answer is $(C).$

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