Answer
$D=95-20\log r$
Work Step by Step
Use the rules for logarithms:
Start with
$\displaystyle \log_{b}\left(\frac{x}{y}\right)=\log_{b}x-\log_{b}y$
We get:
$D=10[\log(320\times 10^{7})-\log r^{2}]$
Apply
$\log_{b}(xy)=\log_{b}x+\log_{b}y$
and
$\log_{b}\left(x^{r}\right)=r\log_{b}x$
Now,
$D=10[\log(320)+\log 10^{7}-2\log r^{2}]$
Apply
$ \log_{b}(b^{x})=x$
$D=10[\log(320)+7-2\log r^{2}]$
$D=25+70-20\log r$
$D=95-20\log r$
