Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 89

$x = 2$

$\ln(x-4) + \ln(x+1) = \ln (x-8)$ $\ln [(x-4)(x+1)] = \ln (x-8)$ $(x-4)(x+1) = (x-8)$ $x(x+1)-4(x+1) = x-8$ $x^{2} + x - 4x - 4 - x + 8 = 0$ $x^{2} - 3x - x - 4 + 8 = 0$ $x^{2} - 4x + 4 = 0$ $(x-2)(x-2) = 0$ $x = 2$