Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.5 Exponential and Logarithmic Equations - 4.5 Exercises - Page 470: 93

$x= -\frac{1}{C}ln(\frac{A+B-y}{B})$

$y=A+B(1-e^{-Cx})\longrightarrow 1-e^{-Cx}=\frac{y-A}{B} \longrightarrow e^{-Cx}=1-\frac{y-A}{B} \longrightarrow -Cx=ln(1-\frac{y-A}{B})\longrightarrow x=-\frac{1}{C}ln(1-\frac{y-A}{B})=-\frac{1}{C}ln(\frac{A+B-y}{B})$