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.
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____
3 answers
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.
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]))
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]))