# How do you solve Ln(7-2x) = Ln(3x+3)  and find any extraneous solutions?

May 20, 2016

$\ln \left(7 - 2 x\right) - \ln \left(3 x + 3\right) = 0$

$\ln \left(\frac{7 - 2 x}{3 x + 3}\right) = 0$

$\frac{7 - 2 x}{3 x + 3} = {e}^{0}$

$\frac{7 - 2 x}{3 x + 3} = 1$

$7 - 2 x = 1 \left(3 x + 3\right)$

$7 - 2 x = 3 x + 3$

$7 - 3 = 5 x$

$4 = 5 x$

$\frac{4}{5} = x$

Hopefully this helps!