Chapter 4 - Exponential and Logarithmic Functions - Section 4.6 Logarithmic and Exponential Equations - 4.6 Assess Your Understanding - Page 338: 83

$ 2,3$

1. $log_9(7x-5)=log_3(x+1) \Longrightarrow \frac{log(7x-5)}{log9}=\frac{log(x+1)}{log3} \Longrightarrow \frac{log(7x-5)}{log3^2}=\frac{log(x+1)}{log3}\Longrightarrow log(7x-5)=2log(x+1) \Longrightarrow (7x-5)=(x+1)^2 \Longrightarrow x^2-5x+6=0 \Longrightarrow x=2,3 $ 2. Check, $x=2,3$ fit.