# How do you condense 2/3ln y + ln t – ln y -1/2 ln t?

Sep 2, 2016

$\frac{2}{3} \ln y + \ln t - \ln y - \frac{1}{2} \ln t = \ln \left(\frac{\sqrt{t}}{\sqrt[3]{y}}\right)$

#### Explanation:

$\frac{2}{3} \ln y + \ln t - \ln y - \frac{1}{2} \ln t$

= $\ln y \left(\frac{2}{3} - 1\right) + \ln t \left(1 - \frac{1}{2}\right)$

= $- \frac{1}{3} \ln y + \frac{1}{2} \ln t$

Now as $\frac{1}{m} \ln a = \ln {a}^{\frac{1}{m}} = \ln \sqrt[m]{a}$ and $- \ln b = \ln \left(\frac{1}{b}\right)$

$- \frac{1}{3} \ln y + \frac{1}{2} \ln t$

= $\ln \left(\frac{1}{\sqrt[3]{y}}\right) + \ln \sqrt{t}$

= $\ln \left(\frac{\sqrt{t}}{\sqrt[3]{y}}\right)$