Chapter 12 - Exponential Functions and Logarithmic Functions - 12.6 Solving Exponential Equations and Logarithmic Equations - 12.6 Exercise Set - Page 825: 51

$x=2$

$\log(2x+1)=\log 5$ First, the solutions must satisfy $2x+1\displaystyle \gt 0\quad \Rightarrow\quad x\gt -\frac{1}{2}\qquad (*)$ in order for the equation to be defined. Apply the principle of logarithmic equality $2x+1=5$ $2x=4$ $x=2\quad $... satisfies (*), and is a valid solution.