## College Algebra (6th Edition)

$x = -7$ Since $x \ne -7$, there is no answer $=$ no solution.
$\log(3x-3) = \log (x+1) + \log 4$ $\log(3x-3) = \log [4(x+1)]$ $\log(3x-3) = \log (4x + 4)$ $\log (3x-3) - \log(4x+4) = 0$ $\log \frac{3x-3}{4x+4} = 0$ $1 = \frac{3x-3}{4x+4}$ $\frac{3x-3}{4x+4} = 1$ $3x - 3 = 1(4x+4)$ $3x - 3 = 4x + 4$ $3x - 4x - 3 - 4 = 0$ $-x - 7 = 0$ $-x = 7$ $x = -7$ Since $x \ne -7$, there is no answer $=$ no solution.