Solve for x: log₃(x + 1) - 1 = 3 + log₃(x - 1)

Given,

log₃(x + 1) - 1 = 3 + log₃(x - 1)

Since log₃3 = 1, the above equation can be written as

⇒ log₃(x + 1) - log₃3 = 3log₃3 + log₃(x - 1)

⇒ log₃(x + 1) - log₃3 = log₃3³ + log₃(x - 1)

⇒ $\text{log}$3\dfrac{(x + 1)}{3} = \text{log}3\space[3^3 \times (x - 1)]

⇒ $\text{log}$3\dfrac{(x + 1)}{3} = \text{log}3\space[27(x - 1)]

⇒ $\dfrac{x + 1}{3} = 27(x - 1)$

⇒ x + 1 = 81(x - 1)

⇒ x + 1 = 81x - 81

⇒ 81x - x = 1 + 81

⇒ 80x = 82

⇒ x = $\dfrac{82}{80} = \dfrac{41}{40} = 1\dfrac{1}{40}.$

Hence, x = $1\dfrac{1}{40}$.