The demand for a new computer game can be modeled by $p(x)=45-5 \ln x$, for $0 \leq x \leq 800$, $R(x)=x \cdot p(x)$. Complete parts (a) through (c) below.
c) How many units will be sold if the price that consumers are willing to pay is $\$ 50$ ?
The number of units that will be soldh s
(Round to the nearest whole number as needed.)

Step 1 ：First, we need to find the value of x that makes the price $50. We can do this by setting the price function equal to 50 and solving for x.

Step 2 ：\(45 - 5 \ln x = 50\)

Step 3 ：Subtract 45 from both sides to isolate the logarithm on one side of the equation.

Step 4 ：\(-5 \ln x = 50 - 45\)

Step 5 ：\(-5 \ln x = 5\)

Step 6 ：Divide both sides by -5 to solve for \(\ln x\).

Step 7 ：\(\ln x = -1\)

Step 8 ：Raise e to the power of both sides to solve for x.

Step 9 ：\(x = e^{-1}\)

Step 10 ：\(x \approx 0.368\)

Step 11 ：However, since the number of units sold must be a whole number, we round 0.368 to the nearest whole number, which is 0.

Step 12 ：So, if the price that consumers are willing to pay is $50, no units will be sold.

Step 13 ：\(\boxed{0}\)