Question

Find the value of ${27}^{\left(\frac{1}{{\log }_{4}3}\right)}+{16}^{{\log }_{4}2}+3^{\left(\frac{4}{{\log }_{7}9}\right)}$

• A. 64
• B. 68
• C. 117
• D. 49

Answer 1

The correct option is C

117

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

$=3^{3{\log }_{3}4}+4^{2{\log }_{4}2}+3^{4{\log }_{3^2}7}$
$\{\frac{1}{{\log }_{b}a}={\log }_{b}a\}$

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

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