Question

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

Answer 1

Given, $a^{lo{g}_3^7}=27$

$a^{log}=a^{(lo{g}_3^7{)}^2}+b^{(lo{g}_7^{11}{)}^2}+c^{(lo{g}_{11}^{25}{)}^2}$

$=a^{lo{g}_3^7\times lo{g}_3^7}+b^{lo{g}_7^{11}\times lo{g}_7^{11}}+c^{lo{g}_{11}^{25}\times lo{g}_{11}^{25}}$

$={27}^{lo{g}_3^7}+{49}^{lo{g}_7^{11}}+(\sqrt{11}{)}^{lo{g}_{11}^{25}}$

$=3^{3lo{g}_3^7}+7^{2lo{g}_7^{11}}+{11}^{\frac{1}{2}lo{g}_{11}^{25}}$

$=7^3+{11}^2+{25}^{\frac{1}{2}}$

$=343+121+5$

$=469$

$\therefore$ Middle value ${=}^{\prime }{6}^{\prime }$

$P\left(E\right)=\frac{1}{2}$.