## Precalculus (6th Edition) Blitzer

Solution set = $\displaystyle \{\frac{1}{5}\}.$
For the equation to be defined, logarithms must have positive arguments $\left[\begin{array}{llll} x+7 \gt 0, & and & 7x+1 \gt 0 & \\ x \gt -7 & & x\gt-1/7 & \end{array}\right]$ $x \gt -1/$7$\qquad (*)$ This is the condition eventual solutions must satisfy. $\log(x+7)-\log 3=\log(7x+1) \qquad$... LHS: quotient rule $\displaystyle \log\frac{x+7}{3}=\log(7x+1) \quad$ ... $\log$ is one-to-one $\displaystyle \frac{x+7}{3}=7x+1 \quad$ ... /$\times$3 $x+7=21x+3$ $7-3=21x-x$ $4=20x$ $x=\displaystyle \frac{4}{20}=\frac{1}{5}\qquad$ ... satisfies (*) Thus, the solution set = $\displaystyle \{\frac{1}{5}\}.$