Question

If $a^{{\log }_{b}c}=3\cdot 3^{{\log }_{4}3}{3}^{{{\log }_{4}3}^{{\log }_{4}3}}{3}^{{{\log }_{4}3}^{{{\log }_{4}3}^{{\log }_{4}3}}}.....\infty$ where $a,b,c$ belongs to set of Rational numbers then the value of $abc=$

• A. $9$
• B. $12$
• C. $16$
• D. $20$

Answer 1

The correct option is C $16$

$a^{{\log }_{b}c}=33^{{\log }_{4}3}{3}^{{{\log }_{4}3}^{{\log }_{4}3}}{3}^{{{\log }_{4}3}^{{{\log }_{4}3}^{{\log }_{4}3}}}.....\infty$
Using the law of exponents of addition,$a^m{a}^n=a^{m+n}$
$=a^{{\log }_{b}c}=3^{1+{\log }_{4}3+{\left({\log }_{4}3\right)}^2+{\left({\log }_{4}3\right)}^3+.....\infty }$
$1+{\log }_{4}3+{\left({\log }_{4}3\right)}^2+{\left({\log }_{4}3\right)}^3+....\infty$ which is in G.P
where $a=1$ and common ratio $r={\log }_{4}3$
As there are infinite terms, $S_{\infty }=\frac{a}{1-r}$
$S_{\infty }=\frac{1}{1-{\log }_{4}3}$
Take $1={\log }_{4}4\Rightarrow \frac{1}{1-{\log }_{4}3}=\frac{1}{{\log }_{4}4-{\log }_{4}3}$
By the quotient law of logarithms, we get
$\frac{1}{{\log }_{4}4-{\log }_{4}3}=\frac{1}{{\log }_4\left(\frac{4}{3}\right)}$
By Base change theorem of logarithms,
$\frac{1}{{\log }_4\left(\frac{4}{3}\right)}={\log }_{\frac{4}{3}}4$
Comparing $a^{{\log }_{b}c}$ with $3^{{\log }_{\frac{4}{3}}4}$
we get $a=3,b=\frac{4}{3},c=4$
Hence, $abc=3\times \frac{4}{3}\times 4=16$