The number of solutions for the equation, ${\left({\log }_{10}x\right)}^2+{\log }_{10}{x}^2={\left({\log }_{10}2\right)}^2-1$  is

${\log }_{10}x=a$

$a^2+2a={\left({\log }_{10}2\right)}^2-1\Rightarrow a^2+2a-{\left({\log }_{10}2\right)}^2+1=0$

$=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-2\pm \sqrt{4-4\left({\left(-{\log }_{10}2\right)}^2+1\right)}}{2}$

$=\frac{-2\pm \sqrt{4+4{\left({\log }_{10}2\right)}^2-4}}{2}$

$=\frac{-2\pm \sqrt{4{\left({\log }_{10}2\right)}^2}}{2}$

$=\frac{-2\pm 2{\log }_{10}2}{2}$

$=\frac{2\left(-1\pm {\log }_{10}2\right)}{2}$

$=-1\pm {\log }_{10}2$

$\therefore$  The number of solutions is 2.