# Thread: Extremely Simple Log Problem =D

1. ## Extremely Simple Log Problem =D

How Would One Go about simplifying this ln(3x^2 - 9x) + ln(1/3x)

2. Hi,
you could use this law of logarithms : $\log_a{(x)} + \log_a{(y)} = \log_a{(xy)}$.

But before proceeding to this, we need a little clarification : is your second term $\ln{\left ( \frac{x}{3} \right )}$ or $\ln{\left (\frac{1}{3x} \right )}$ ?

3. $

\ln{\left (\frac{1}{3x} \right )}
$

It Looks like this

4. Then, using the law of logarithms I mentioned in my previous post, you have :

$\ln{\left ( 3x^2 - 9x \right )} + \ln{\left ( \frac{1}{3x} \right )} = \ln{\left [ \left ( 3x^2 - 9x \right ) \left ( \frac{1}{3x} \right ) \right ]} = \ln{\left ( \frac{3x^2 - 9x}{3x} \right )}$

Note that the top part of the fraction, $3x^2 - 9x$, can be factored as $3x(3x - 3)$. Therefore :

$\ln{\left ( \frac{3x^2 - 9x}{3x} \right )} = \ln{\left ( \frac{3x(3x - 3)}{3x} \right )} = \ln{\left ( 3x - 3 \right )}$

Conclusion : $\ln{\left ( 3x^2 - 9x \right )} + \ln{\left ( \frac{1}{3x} \right )} = \ln{\left ( 3x - 3 \right )}$

Does this make sense ?

5. Thank You very much very helpful