Question

If the value of the product $P=3\cdot 3^{{\log }_{4}3}\cdot 3^{{\log }_4{3}^{{\log }_{4}3}}\cdot 3^{{\log }_4{3}^{{\log }_4{3}^{{\log }_{4}3}}}\dots \dots \dots$ 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)

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