$\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{(2x-2).2x}=\dfrac{1}{8}$
$\dfrac{1}{2}.(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x})=\dfrac{1}{8}$
$\dfrac{1}{2}.(\dfrac{1}{2}-\dfrac{1}{2x})=\dfrac{1}{8}$
$\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}$
$\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}$
$⇒2x=4$
$⇒x=\dfrac{4}{2}=2$