## Precalculus (6th Edition) Blitzer

Solution set = $\displaystyle \{\frac{11}{3}\}.$
For the equation to be defined, logarithms must have positive arguments $\left[\begin{array}{lll} x-2 \gt 0, & \Rightarrow & x \gt 2\\ x+3\gt 0, & \Rightarrow & x \gt -3\\ x-1 \gt 0, & \Rightarrow & x \gt 1\\ x+7 \gt 0, & \Rightarrow & x \gt -7 \end{array}\right]$ $x \gt 2 \qquad (*)$ This is the condition eventual solutions must satisfy. $\ln(x-2)-\ln(x+3)=\ln(x-1)-\ln(x+7)$ We use the quotient rule on both sides: $\ln \displaystyle \frac{x-2}{x+3}=\ln\frac{x-1}{x+7}$ Recall that ln is one to one: $\displaystyle \frac{x-2}{x+3}=\frac{x-1}{x+7}$ Multiply with the LCD, $(x+3)(x+7)$: $(x-2)(x+7)=(x+3)(x-1)$ $x^{2}+5x-14=x^{2}+2x-3$ $5x-2x=-3+14$ $3x=11$ $x=\displaystyle \frac{11}{3} \qquad$ ... satisfies (*) Thus, the solution set = $\displaystyle \{\frac{11}{3}\}.$