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CAT 2018 Set-1 | Question: 71

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1 Answer

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Given that,

  • $ \log_{2} (5+ \log_{3} {a}) = 3 \quad \longrightarrow (1)$
  • $ \log_{5} ( 4a + 12) = 3 \quad \longrightarrow (2)$

From equation $(1),$

$\log_{2} (5 + \log_{3}{a}) = 3$

$ \Rightarrow 5 + \log_{3}{a} = 2^{3}$ $ \quad [ \because {\log_{a}{b}} = x \Rightarrow b = a^{x}]$

$ \Rightarrow \log_{3}{a} = 3$

$ \Rightarrow a = 3^{3}$

$ \Rightarrow \boxed {a = 27} $

From equation $(2),$

$ \log_{5} (4a + 12 + \log_{2}{b}) = 3 $

$ \Rightarrow 4a + 12 +\log_{2}{b} = 5^{3}$

$ \Rightarrow 4(27) + 12 + \log_{2}{b} = 125 $

$ \Rightarrow 108 + 12 + \log_{2}{b} = 125 $

$ \Rightarrow 120 + \log_{2}{b} = 125 $

$ \Rightarrow \log_{2}{b} = 125 – 120 $

$ \Rightarrow \log_{2}{b} = 5 $

$ \Rightarrow b = 2^{5}$

$ \Rightarrow \boxed {b = 32} $

$ \therefore$ The value of $ a+b = 27 + 32 = 59 $

Correct Answer $: \text {D}$

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