Answer :
Recall that:
[tex]\log _ab=\log _ac\Leftrightarrow b=c.[/tex]Therefore:
[tex]x^2+11=15.[/tex]Subtracting 11 from the above equation we get:
[tex]\begin{gathered} x^2+11-11=15-11, \\ x^2=4. \end{gathered}[/tex]Therefore:
[tex]x=\pm2.[/tex]Answer: Second option.