## Precalculus (6th Edition) Blitzer

Solution set = $\emptyset.$
For the equation to be defined, logarithms must have positive arguments $\left[\begin{array}{lll} x-5 \gt 0, & \Rightarrow & x \gt 5\\ x+4\gt 0, & \Rightarrow & x \gt -4\\ x-1 \gt 0, & \Rightarrow & x \gt 1\\ x+2 \gt 0, & \Rightarrow & x \gt -2 \end{array}\right]$ $x \gt 5 \qquad (*)$ This is the condition eventual solutions must satisfy. $\ln(x-5)-\ln(x+4)=\ln(x-1)-\ln(x+2)$ Apply the quotient rule on both sides: $\displaystyle \ln\frac{x-5}{x+4}=\ln\frac{x-1}{x+2}$ Recall that ln is one to one: $\displaystyle \frac{x-5}{x+4}=\frac{x-1}{x+2}$ Multiply with the LCD, $(x+4)(x+2)$: $(x-5)(x+2)=(x+4)(x-1)$ $x^{2}-3x-10=x^{2}+3x-4$ $-3x-3x=-4+10$ $-6x=6$ $x=-1 \quad$ ... does not satisfy (*) Thus, the solution set = $\emptyset.$