The demand equation for a hand-held electronic organizer is... P=5000(1-(4/4+e^-.002x))
Find the demand x for a price of (a) P=600?
So strange. Thx for help
ab
correct answers: -303.1 or 500ln(6/11)
To find the value of x when the demand equation is equal to a specific price, you need to substitute the given price value into the equation and solve for x.
The given demand equation is:
P = 5000(1-(4/4+e^(-0.002x)))
To find the demand x for a price of P = 600, you need to substitute P = 600 into the equation:
600 = 5000(1-(4/4+e^(-0.002x)))
Now, let's solve for x:
Divide both sides of the equation by 5000:
(1-(4/4+e^(-0.002x))) = 0.12
Subtract 1 from both sides:
-((4/4+e^(-0.002x))) = -0.88
Multiply both sides by -1:
4/4 + e^(-0.002x) = 0.88
Subtract 4/4 from both sides:
e^(-0.002x) = 0.88 - 1
e^(-0.002x) = -0.12
To isolate e^(-0.002x), take the natural logarithm (ln) of both sides:
ln[e^(-0.002x)] = ln(-0.12)
-0.002x = ln(-0.12)
Now, divide both sides by -0.002:
x = ln(-0.12) / -0.002
At this point, the logarithm of a negative number is undefined, so it seems there may be an error in the given demand equation or the provided price value. Double-check your calculations or ensure that the values are correct.
Don't you mean
P = 5000[1 - 4/(4 + e^-.002x)]
?
If P = 600 ,
0.120 = [1 - 4/(4 + e^-.002x)]
4/(4 + e^-.002x)] = 0.88
4 + e^-.002x = 4.5454
e^-.002x = 0.5454
-0.002x = ln(0.5454) = -0.6061
x = 303