Question

If the possible real values of $x$ for which the sixth term of ${(3^{{\log }_3\sqrt{9^{|x-2|}}}+7^{\left(\frac{1}{5}\right){\log }_7{3}^{|x-2|-9}})}^7$ equal $567$ are $x_1$ and $x_2$ then value of $\frac{x_1^2+x_2^2}{3}$

Answer 1

${=}^7{C}_5{(3^{{\log }_3\sqrt{9^{|x-2|}}}+7^{\left(\frac{1}{5}\right){\log }_7{3}^{|x-2|-9}})}^7=567$

$\Rightarrow 9^{|x-2|}\times 3^{|x-2|-9}=562$

$\Rightarrow 9^{|x-2|}\times 3^{|x-2|-9}=\frac{567}{21}=27$

$3^{2|x-2|}\times 3^{|x-2|}\times 3^{-9}=27$

$3^{3|x-2|}=3^3\times 3^9$

$(3^3{)}^{(x-2)}=(3^3{)}^4$

$\therefore |x-2|=4$

$\therefore x-2=4$

$\therefore x=6$

$x-2=-4$

$x=-2$

Let $x_1=6,x_2=-2$

$\therefore \frac{x_1^2+x_2^2}{3}=\frac{(6{)}^2+(-2{)}^2}{3}$

$=\frac{36+4}{3}=\frac{40}{3}$.