The demand function for a product is given by

p = 10,000 [1 − (5/5 + e^−0.001x)]

where p is the price per unit (in dollars) and x is the number of units sold. Find the numbers of units sold for prices of
p = $1000
and
p = $1500.
(Round your answers to the nearest integer.)

(a) p = $1000 _____units
(b) p = $1500 _____units

To find the number of units sold for a given price, we need to solve the demand function for x when p is equal to the given price.

(a) For p = $1000:
Substitute p = $1000 into the demand function and solve for x:

$1000 = 10,000 [1 − (5/5 + e^−0.001x)]

Dividing both sides of the equation by 10,000:

0.1 = 1 − (5/5 + e^−0.001x)

Using the property a/(b+c) = a/b + a/c:

0.1 = 1 − (1 + e^−0.001x)

Combining like terms:

0.1 = 1 - 1 - e^−0.001x

0.1 = - e^−0.001x

Taking the natural logarithm of both sides:

ln(0.1) = ln(− e^−0.001x)

Using the property ln(e^a) = a:

ln(0.1) = −0.001x

Solving for x:

x = ln(0.1) / −0.001

Using a calculator, we find:

x ≈ 2296

Hence, for p = $1000, approximately 2296 units are sold.

(b) For p = $1500:
Follow the same steps as above, but substitute p = $1500 into the demand function:

$1500 = 10,000 [1 − (5/5 + e^−0.001x)]

Dividing both sides by 10,000:

0.15 = 1 − (5/5 + e^−0.001x)

Using the property a/(b+c) = a/b + a/c:

0.15 = 1 − (1 + e^−0.001x)

Combining like terms:

0.15 = 1 - 1 - e^−0.001x

0.15 = - e^−0.001x

Taking the natural logarithm of both sides:

ln(0.15) = ln(− e^−0.001x)

Using the property ln(e^a) = a:

ln(0.15) = −0.001x

Solving for x:

x = ln(0.15) / −0.001

Using a calculator, we find:

x ≈ 3470

Hence, for p = $1500, approximately 3470 units are sold.