Solve for y in the equation: 5^(2y + 1) = 4(5)^(y + 1) – 15

Rewrite using u = 5^y. Then 5^(2y + 1) = 5u² and 4(5)^(y + 1) = 20u.

The equation becomes 5u² = 20u – 15. Divide by 5: u² – 4u + 3 = 0, so (u – 1)(u – 3) = 0 and u = 1 or u = 3.

If 5^y = 1, then y = 0. If 5^y = 3, then y = log₅3. So y = log₅3 or y = 0.