The demand for a new computer game can be modeled by $p(x)=52\mathrm{.}5-5lnx,$ for $0x800,$ where $p(x)$ is the price consumers will pay, in dollars, and $x$ is the number of games sold, in thousands. Recall that total revenue is given by $R(x)=x*p(x).$ Complete parts $($a$)$ through $($c$)$ below.

a$)$ Find $R(x).$

$R(x)=$

b$)$ Find the marginal revenue, $R^{\text{'}}(x).$

$R^{\text{'}}(x)=47\mathrm{.}5-5ln(x)$

c$)$ How many units will be sold if the price that consumers are willing to pay is $$40?$

The number of units that will be sold is

$($Round to the nearest whole number as needed.$)$