Correct Answer: Option C

Explanation

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 = \(\frac{2}{3}\)

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