Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 88

Since $x \ne -5$, then $x = 4$

$\log (x+3) + \log (x-2) = \log 14$ $\log [(x+3)(x-2)] = \log 14$ $(x+3)(x-2) = 14$ $x(x-2)+3(x-2) = 14$ $x^{2} - 2x + 3x - 6 - 14 = 0$ $x^{2} + x - 20 = 0$ $(x+5)(x-4) = 0$ $x = -5, 4$ Since $x \ne -5$, then $x = 4$