How Would One Go about simplifying this ln(3x^2 - 9x) + ln(1/3x)
Hi,
you could use this law of logarithms : $\displaystyle \log_a{(x)} + \log_a{(y)} = \log_a{(xy)}$.
But before proceeding to this, we need a little clarification : is your second term $\displaystyle \ln{\left ( \frac{x}{3} \right )}$ or $\displaystyle \ln{\left (\frac{1}{3x} \right )}$ ?
Then, using the law of logarithms I mentioned in my previous post, you have :
$\displaystyle \ln{\left ( 3x^2 - 9x \right )} + \ln{\left ( \frac{1}{3x} \right )} = \ln{\left [ \left ( 3x^2 - 9x \right ) \left ( \frac{1}{3x} \right ) \right ]} = \ln{\left ( \frac{3x^2 - 9x}{3x} \right )}$
Note that the top part of the fraction, $\displaystyle 3x^2 - 9x$, can be factored as $\displaystyle 3x(3x - 3)$. Therefore :
$\displaystyle \ln{\left ( \frac{3x^2 - 9x}{3x} \right )} = \ln{\left ( \frac{3x(3x - 3)}{3x} \right )} = \ln{\left ( 3x - 3 \right )}$
Conclusion : $\displaystyle \ln{\left ( 3x^2 - 9x \right )} + \ln{\left ( \frac{1}{3x} \right )} = \ln{\left ( 3x - 3 \right )}$
Does this make sense ?