Question

If a, b, c are positive real numbers such that $a^{lo{g}_{3}7}=27;b^{lo{g}_{7}11}=49$ and $c^{lo{g}_{11}25}=\sqrt{11}$. The value of $(a^{(lo{g}_{3}7{)}^2}+b^{(lo{g}_{7}11{)}^2}+c^{(lo{g}_{11}25{)}^2})$ equals

• A. $489$
• B. $469$
• C. $464$
• D. $400$

Answer 1

Answer: (B)

$469$

$a^{(lo{g}_{3}7{)}^2}=a^{(lo{g}_{3}7\times lo{g}_{3}7)}$

$=(a^{lo{g}_{3}7}{)}^{lo{g}_{3}7}$

$=(27{)}^{lo{g}_{3}7}$

similarly,

$b^{(lo{g}_{7}11{)}^2}=b^{(lo{g}_{7}11\times lo{g}_{7}11)}$

$=(49{)}^{lo{g}_{7}11}$

$c^{(lo{g}_{11}25{)}^2}=c^{(lo{g}_{11}25\times lo{g}_{11}25)}$

$=(\sqrt{11}{)}^{lo{g}_{11}25}$

Now,

$\left(a^{(lo{g}_{3}7{)}^2}\right)+\left(b^{(lo{g}_{7}11{)}^2}\right)+\left(c^{(lo{g}_{11}25{)}^2}\right)$

$=(27{)}^{lo{g}_{3}7}+(49{)}^{lo{g}_{7}11}+(\sqrt{11}{)}^{lo{g}_{11}25}$

$=(3{)}^{3lo{g}_{3}7}+(7{)}^{2lo{g}_{7}11}+(11{)}^{\frac{1}{2}lo{g}_{11}25}$

$=7^3+{11}^2+(25{)}^{\frac{1}{2}}$...........{$\because a^{lo{g}_{a}x}=x$}

$=343+121+5=469$