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