1. ## Exponential problem

Hi,

I'm stuck in this problem. Can somebody help me?

After discontinuing all advertising for a certain product in 1994 ,the manufacturer noted that sales began to drop according to the model

s = 500,000/(1+.6e^(kt))

where s represents the number of units sold and t = 0 represents 1994. In 1996, the company sold 300,000 units

a) complete the model by solving for k
b) estimates sales in 1999

Thank you ...(really!)

Isabel

2. Originally Posted by cuteisa89
After discontinuing all advertising for a certain product in 1994 ,the manufacturer noted that sales began to drop according to the model

s = 500,000/(1+.6e^(kt))

where s represents the number of units sold and t = 0 represents 1994. In 1996, the company sold 300,000 units

a) complete the model by solving for k
i suppose t is the time in years.

in 1996, the company sold 300,000 units, that means when t = 2, s = 300,000. now we can use this to solve for k

=> 300000 = 500000/(1 + 0.6e^(2k))
=> 300000(1 + 0.6e^(2k)) = 500000
=> 1 + 0.6e^(2k) = 500000/300000 = 5/3
=> 0.6e^(2k) = 5/3 - 1 = 2/3
=> e^(2k) = 2/3*10/6 = 10/9 ...................0.6 = 6/10 so i multiplied both sides by 10/6

now we take the log of both sides

=> lne^(2k) = ln(10/9)
=> 2k*ln(e) = ln(10/9)
=> 2k = ln(10/9) ........................lne = 1
=> k = ln(10/9)/2 = 0.0527

so the complete model is s = 500,000/(1 + 0.6e^(0.0527t))

b) estimates sales in 1999
in 1999, t = 5

=> s = 500,000/(1 + 0.6e^(0.0527(5)))
=> s = 500000/(1 + 0.7809)
=> s = 280756.92 units