Question

If $a,b,c$ are positive real numbers such that $a^{{\log }_{3}7}=27,b^{{\log }_{11}7}=49\text{and}{c}^{{\log }_{11}25}=\sqrt{11}$, then the middle digit in the value of $(a^{({\log }_{3}7{)}^2}+b^{({\log }_{7}11{)}^2}+c^{({\log }_{11}25{)}^2})$ equals to

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is D $6$

$a^{({\log }_{3}7{)}^2}={\left(a^{{\log }_{3}7}\right)}^{{\log }_{3}7}=(27{)}^{{\log }_{3}7}=7^3=343$

$b^{({\log }_{7}11{)}^2}=(49{)}^{{\log }_{7}11}={11}^2=121$
$c^{({\log }_{11}25{)}^2}=(\sqrt{11}{)}^{{\log }_{11}25}=(25{)}^{1/2}=5$

$\therefore \text{Sum}=343+121+5=469$
$\Rightarrow \text{Middle digit}=6$