## College Algebra (6th Edition)

NO SOLUTION as $x \ne - 10$
$\log(2x-1) = \log(x+3) + \log 3$ $\log (2x-1) = \log [3(x+3)]$ $\log (2x-1) = \log (3x + 9)$ $\log(2x-1)-\log(3x+9) = 0$ $\log \frac{2x-1}{3x+9} = 0$ $1 = \frac{2x-1}{3x+9}$ $\frac{2x-1}{3x+9} = 1$ $2x - 1 = 1(3x+9)$ $2x - 1 = 3x + 9$ $2x - 3x - 1 - 9 = 0$ $-x - 10 = 0$ $-x = 10$ $x = -10$ NO SOLUTION as $x \ne - 10$