SOLVE THE BELOW LOGARITHMIC EQUATION

Question

 It is given that (2023) = (2032)^X+1.   Show that X =1/[log₂₀₃₂ 2023] − 1

Answer 1

2023 = (2032)^x+1

⇒ log₂₀₃₂ 2023 = x+1 (By taking log₂₀₃₂ both sides)

⇒ x = log₂₀₃₂ 2023 - 1

⇒ x = \(\frac{1}{log_{2023}2032}\) - 1 (∵ log_a b = \(\frac{1}{log_ba}\))