Question

If $x_1,x_2,$ $x_3$ are the three real solutions of the equation;
$x^{{\log }_{10}2x+{\log }_{10}{x}^3+3}=\frac{2}{\frac{1}{\sqrt{x+1}-1}-\frac{1}{\sqrt{x+1}+1}}where{x}_1>x_2>x_3,$ then

• A. $x_1+x_3=2x_2$
• B. $x_1+x_3=x_2^2$
• C. $x_2=-\frac{2x_1{x}_2}{x_1+x_2}$
• D. $x{{}_1}^{-1}+x{{}_2}^{-1}=x{{}_3}^{-1}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $x_1+x_3=2x_2$
B $x_1+x_3=x_2^2$
C $x_2=-\frac{2x_1{x}_2}{x_1+x_2}$

$x^{{\log }_{10}\left(2x{x}^3\right)+3}=\frac{2\left(\right(x+1)-1)}{(\sqrt{x+1}+1)-(\sqrt{x+1}-1)}{x}^{{\log }_{10}\left(2x^4\right)+3}=x{\log }_{10}\left(2x^4\right)+3=12x^4={10}^{-2}{x}^4=\frac{1}{200}x=\pm \frac{1}{{\left(200\right)}^{1/4}},\pm \frac{i}{{\left(200\right)}^{1/4}}{x}_3=-\frac{1}{{\left(200\right)}^{1/4}},x_2=0,x_1=\frac{1}{{\left(200\right)}^{1/4}}\left(A\right)x_1+x_3=2x_2-\frac{1}{{\left(200\right)}^{1/4}}+\frac{1}{{\left(200\right)}^{1/4}}=00=0\left(B\right)x_1+x_3=x_2^2-\frac{1}{{\left(200\right)}^{1/4}}+\frac{1}{{\left(200\right)}^{1/4}}=0^2=0\left(C\right)x_2=-\frac{2x_1{x}_2}{(x_1+x_2)}0=-\frac{2\frac{1}{{\left(200\right)}^{1/4}}\left(0\right)}{\frac{1}{{\left(200\right)}^{1/4}}+0}0=0\left(D\right)\frac{1}{x_1}+\frac{1}{x_2}=\frac{1}{x_3}{\left(200\right)}^{-1/4}+\frac{1}{0}=-{\left(200\right)}^{-1/4}\infty \ne -{\left(200\right)}^{-1/4}$