1. ## natural log

solve the equation for x:

ln x + ln (x-3) = 0

2. Originally Posted by algebra2
solve the equation for x:

ln x + ln (x-3) = 0
use the product property of logs ... $\ln{a} + \ln{b} = \ln(ab)$

$\ln[x(x-3)] = 0$

change to an exponential equation ...

$x(x-3) = e^0$

$x(x-3) = 1$

solve the quadratic equation ... remember that only values of x > 3 are valid.

3. ln(x) +l(x-3) =0

ln[x(x-3)] =0

x(x-3) = e^0

x(x-3)=1

x^2 - 3x -1=0 (Use the quadratic formula to find the roots)