10\(^y\) x 5\(^{(2y - 2)}\) x 4\(^{(y - 1)}\) = 1 but 10\(^y\) = (5 x 2)\(^y\) = 5\(^y\) x 2\(^y\) = (Law of indices) 5\(^y\) x 2\(^y\) x 5\(^{(2y - 2)}\) x 4\(^{(y - 1)}\) = 1 but 4\(^{(y - 1)}\) = 2\(^{2(y - 1)}\) = 2\(^{2y - 2}\) (Law of indices) 5\(^y\) x 5\(^{(2y -2)}\) x 2\(^{(2y- 2)}\) x 2\(^y\) = 1 5\(^{(3y -2)}\) x 2\(^y\) x 2\(^{(2y -2)}\) = 1 = 5\(^{(3y -2)}\) x 2\(^{(3y -2)}\) = 1 But a\(^o\) = 1 10\(^{(3y -2)}\) = 10\(^o\) 3y - 2 = 0 ∴ y = \(\frac{2}{3}\)