\( ( 1 - \frac { 1 } { 7 } ) \cdot ( 1 - \frac { 1 } { 8 } ) \cdot ( 1 - \frac { 1 } { 9 } ) \cdot \cdots \cdot ( 1 - \frac { 1 } { x } ) = \frac { 1 } { 6 } \)

To solve the equation \( ( 1 - \frac { 1 } { 7 } ) \cdot ( 1 - \frac { 1 } { 8 } ) \cdot ( 1 - \frac { 1 } { 9 } ) \cdots ( 1 - \frac { 1 } { x } ) = \frac { 1 } { 6 } \), we first simplify each term in the product. Each term can be rewritten as: \[ 1 - \frac{1}{n} = \frac{n-1}{n} \] Thus, the product becomes: \[ \frac{6}{7} \cdot \frac{7}{8} \cdot \frac{8}{9} \cdots \frac{x-1}{x} \] Observing the pattern, we can see that all the intermediate terms will cancel out, resulting in: \[ \frac{6}{x} \] Setting this equal to \( \frac{1}{6} \): \[ \frac{6}{x} = \frac{1}{6} \] Cross-multiplying gives us: \[ 6 \cdot 6 = x \implies x = 36 \] Thus, the solution to the equation is: \[ \boxed{36} \]