Question

If $x={\log }_{2}3,y=1-{\log }_{1331}343$ and $z={\log }_7\left(\frac{1}{3}\right),$ then the value of $9^{1/x}+{121}^y+{\left(\frac{1}{7}\right)}^z$ is

• A. $\frac{463}{49}$
• B. $\frac{354}{49}$
• C. $\frac{170}{49}$
• D. $\frac{464}{49}$

Answer 1

The correct option is D $\frac{464}{49}$

$y=1-{\log }_{1331}343=1-{\log }_{{11}^3}\left(7^3\right)={\log }_{11}11-{\log }_{11}7={\log }_{11}\left(\frac{11}{7}\right)$

$x={\log }_{2}3\Rightarrow \frac{1}{x}={\log }_{3}2$

Now, $9^{1/x}+{121}^y+{\left(\frac{1}{7}\right)}^z$
$=3^{2\times {\log }_{3}2}+{11}^{2\times {\log }_{11}\left(\frac{11}{7}\right)}+7^{-{\log }_7\left(\frac{1}{3}\right)}=3^{{\log }_{3}4}+{11}^{{\log }_{11}{\left(\frac{11}{7}\right)}^2}+7^{{\log }_{7}3}=4+{\left(\frac{11}{7}\right)}^2+3[\because a^{{\log }_{a}m}=m]=7+{\left(\frac{11}{7}\right)}^2=\frac{464}{49}$