Basic loggin' 2

$\\ A=\frac { 1 }{ \log _{ 3 }{ 2 } \times \log _{ 4 }{ 3 } \times \log _{ 5 }{ 4 } \times \log _{ 6 }{ 5 } \times \log _{ 7 }{ 6 } \times \log _{ x }{ 7 } } \\ \\ From\quad \log _{ 2 }{ x } =3,\quad we\quad obtain\quad x=8.\\ \\ Plug\quad x=8\quad into\quad A:\\ A=\frac { 1 }{ \log _{ 3 }{ 2 } \times \log _{ 4 }{ 3 } \times \log _{ 5 }{ 4 } \times \log _{ 6 }{ 5 } \times \log _{ 7 }{ 6 } \times \log _{ 8 }{ 7 } } \\ \\ Observe\quad that\quad \log _{ b }{ a } \times \log _{ c }{ b=\log _{ b }{ a } \times \frac { 1 }{ \log _{ b }{ c } } = } \frac { \log _{ b }{ a } }{ \log _{ b }{ c } } =\log _{ c }{ a } .\\ As\quad for\quad our\quad problem:\quad \log _{ 7 }{ 6 } \times \log _{ 8 }{ 7 } =\log _{ 8 }{ 6 } \\ \qquad \qquad \qquad \qquad \qquad \qquad \log _{ 6 }{ 5 } \times \log _{ 8 }{ 6 } =\log _{ 8 }{ 5 } \\ \qquad \qquad \qquad \qquad \qquad \qquad \log _{ 5 }{ 4 } \times \log _{ 8 }{ 5 } =\log _{ 8 }{ 4 } \\ This\quad goes\quad on\quad until\quad we\quad arrive\quad at:\quad A=\frac { 1 }{ \log _{ 3 }{ 2 } \times \log _{ 8 }{ 3 } } =\frac { 1 }{ \log _{ 8 }{ 2 } } =\frac { 1 }{ \frac { 1 }{ 3 } } \\ So\quad the\quad answer\quad is\quad \boxed { A=3 } .$