Question

If the value of the product $P =3 \cdot 3^{\log _4 3} \cdot 3^{\log _4 3^{\log _4 3}} \cdot 3^{\log _4 3^{\log _4 3^{\log _4 3}}} \ldots \ldots \ldots$ is $a ^{\log _b c }$ where $a , b , c \in Q$, then $b$ equals

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{3}{2}$
• D. $\frac{2}{3}$

Answer 1

Answer: (B)

$P=3 \cdot 3^{\log _4 3} \cdot 3^{\log _4 3^{\log _4 3}} \cdot 3^{\log _4 3^{\log _4 3^{\log _4 3}}} \ldots \ldots \infty$
$\log _3 P =1+\log _4 3+\left(\log _4 3\right)^2+\ldots \ldots \ldots+\infty=\frac{1}{1-\log _4 3} $
$\therefore P=(3)^{\frac{1}{1-\log _4 3}}=(3)^{\frac{1}{\log _4(4 / 3)}}=(3)^{\log _{4 / 3} 4}=a^{\log _b c}$
Hence $b=\frac{4}{3} A$