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____
(b) p = $1500____

what's the problem? Set p=1000 and solve for x:

1000 = 10,000(1 − (5/5 + e^−0.001x))
0.1 = 1 − (5/5 + e^−0.001x)
0.9 = 5/5 + e^−0.001x
I suspect a typo,for 5/5 makes little sense. However, moving right along,
-0.1 = e^-.001x
Now we know there's a boo-boo, since the exponential function is never negative.

I suggest you fix the error, then continue on with the solution. When you get to the point we reached above, take log of both sides.

p = 10,000(1 − (5/5 + e^−0.001x))

for p=1000,
1000=10000(1 − (5/5 + e^−0.001x))
5/(5+e^(-0.001x)=0.9
e^(-0.001x)=5/0.9-5
take log
-0.001x = log(5/0.9-5)
x=-1000 log(5/0.9-5)
=587.8,
say 588

(b) can be done in a similar way.

Steve is right.

Whenever you transpose a fraction from a book to a single line to post, you need to insert the implicit parentheses. In this case, I believe it is
1000 = 10,000(1 − (5/[5 + e^−0.001x]))

To find the number of units sold for prices of p = $1000 and p = $1500, we need to solve the demand function equation for x.

Given: p = 10,000(1 − (5/5 + e^−0.001x))

(a) For p = $1000:
Substitute p = 1000 into the demand function equation:
1000 = 10,000(1 − (5/5 + e^−0.001x))

Now we can solve for x. Let's break down the steps:

1. Divide both sides by 10,000:
1000/10,000 = 1 − (5/5 + e^−0.001x)

2. Simplify the right side:
0.1 = 1 − (1 + e^−0.001x)

3. Combine like terms on the right side:
0.1 = -e^−0.001x

4. Multiply both sides by -1:
-0.1 = e^−0.001x

5. Take the natural logarithm of both sides:
ln(-0.1) = ln(e^−0.001x)

6. Use the logarithmic property to bring down the exponent:
ln(-0.1) = -0.001x

7. Solve for x:
x ≈ ln(-0.1) / -0.001

Perform the calculations on a calculator and round the result to the nearest integer to find the value of x for p = $1000.

(b) For p = $1500:
Repeat the same steps as above, but substitute p = 1500 and solve for x. Round the result to the nearest integer to find the value of x for p = $1500.