For a real number $a,$ if $\dfrac{\log_{15}a + \log_{32}a}{(\log_{15}a)(\log_{32}a)} = 4$ then $a$ must lie in the range

- $a>5$
- $3<a<4$
- $4<a<5$
- $2<a<3$

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1
votes

For a real number $a,$ if $\dfrac{\log_{15}a + \log_{32}a}{(\log_{15}a)(\log_{32}a)} = 4$ then $a$ must lie in the range

- $a>5$
- $3<a<4$
- $4<a<5$
- $2<a<3$

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1
votes

Given that, $\dfrac{\log_{15}{a} + \log_{32}{a}}{(\log_{15}{a})(\log_{32}{a})} = 4$

$\Rightarrow \log_{15}{a} + \log_{32}{a} = 4\left[(\log_{15}{a})(\log_{32}{a})\right]$

$\Rightarrow \dfrac{\log_{c}{a}}{\log_{c}{15}} = 4 \times \dfrac{\log_{c}{a}}{\log_{c}{15}} \times \dfrac{\log_{c}{a}}{\log_{c}{32}}$

$\Rightarrow \require{cancel} \cancel{\log_{c}{a}} \left[\dfrac{\log_{c}{32} + \log_{c}{15}}{\cancel{(\log_{c}{15})} \cdot \cancel{(\log_{c}{32})}}\right] = 4 \times \dfrac{\cancel{\log_{c}{a}}}{\cancel{\log_{c}{15}}} \times \dfrac{\log_{c}{a}}{\cancel{\log_{c}{32}}}$

$\Rightarrow \log_{c}{32} + \log_{c}{15} = 4\log_{c}{a}$

$\Rightarrow \log_{c}{480} = \log_{c}{a}^{4}$

$\Rightarrow \boxed{a^{4} = 480}$

We know that,

- $4^{4} = 256$
- $5^{4} = 625$

$\therefore \boxed{4<a<5}$

Correct Answer $:\text{C}$

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