Question

The value of ${27}^{\left(\frac{1}{{\log }_{4}3}\right)}+{16}^{{\log }_{4}2}+3^{\left(\frac{4}{{\log }_{7}9}\right)}$ is equal to

• A. $64$
• B. $68$
• C. $49$
• D. $117$

Answer 1

The correct option is D

$117$

Given:
${27}^{\left(\frac{1}{{\log }_{4}3}\right)}+{16}^{{\log }_{4}2}+3^{\left(\frac{4}{{\log }_{7}9}\right)}$

$=3^{\frac{3}{{\log }_{4}3}}+4^{2{\log }_{4}2}+3^{\left(\frac{4}{{\log }_7{3}^2}\right)}$


$=3^{3{\log }_{3}4}+4^{2{\log }_{4}2}+3^{4{\log }_{3^2}7}$
$\{\frac{1}{{\log }_{b}a}={\log }_{b}a\}$
$[{\log }_{a^c}b=\frac{1}{c}{\log }_{a}b]$
$=3^{{\log }_3{4}^3}+4^{{\log }_4{2}^2}+3^{\left(\frac{4}{2}\right){\log }_{3}7}$

$=3^{{\log }_3{4}^3}+4^{{\log }_4{2}^2}+3^{{\log }_3{7}^2}$
As
$a^{{\log }_a\left(b\right)}=b$

So,
$=4^3+2^2+7^2$
$=64+4+49=117$