## College Algebra (6th Edition)

Since $x \ne -5$, then $x = 2$
$\log x + \log (x+3) = \log 10$ $\log [x(x+3)] = \log 10$ $\log (x^{2} + 3x) = \log 10$ $x^{2} + 3x = 10$ $x^{2} + 3x - 10 = 0$ $(x+5)(x-2) = 0$ $x = -5, 2$ Since $x \ne -5$, then $x = 2$