Find ln(r), where  $r = \sqrt[3]{\frac{3 V}{4 π}}$  

To find $\ln(r)$ given $r = \sqrt[3]{\frac{3V}{4\pi}}$, we'll start by taking the natural logarithm of both sides of the equation.

$r = \sqrt[3]{\frac{3V}{4\pi}}$

$\ln(r) = \ln\left(\sqrt[3]{\frac{3V}{4\pi}}\right)$

Now, using the property of logarithms that $\ln(a^b) = b \ln(a)$, we can get,

$\ln(r) = \frac{1}{3} \ln\left(\frac{3V}{4\pi}\right)$

Since $\ln(\frac{a}{b}) = \ln(a) - \ln(b)$, we can further simplify:

$\ln(r) = \frac{1}{3} \left(\ln(3V) - \ln(4\pi)\right)$

Now, using the property $\ln(ab) = \ln(a) + \ln(b)$, we simplify $\ln(4\pi)$:

$\ln(r) = \frac{1}{3} \left(\ln(3V) - \ln(4) - \ln(\pi)\right)$

$\ln(r) = \frac{1}{3} \left(\ln(3V) - \ln(4) - \ln(\pi)\right)$

$\ln(r) = \frac{1}{3} \left(\ln(3V) - \ln(4) - \ln(\pi)\right)$

$\ln(r) = \frac{1}{3} \left(\ln(3V) - \ln(4) - \ln(\pi)\right)$

$\ln(r) = \frac{1}{3} \left(\ln(3) + \ln(V) - \ln(4) - \ln(\pi)\right)$

$\ln(r) = \frac{1}{3} \left(\ln(3) + \ln(V) - \ln(4) - \ln(\pi)\right)$

So, $\ln(r) = \frac{1}{3} \left(\ln(3) + \ln(V) - \ln(4) - \ln(\pi)\right)$.