SOLUTION: Please help me with this word problem: A person buys a cup of coffee and drinks it in the store. The coffee's temperature y(in degrees Fahrenheit) is given by y=70+137e^kt, where t

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Please help me with this word problem: A person buys a cup of coffee and drinks it in the store. The coffee's temperature y(in degrees Fahrenheit) is given by y=70+137e^kt, where t      Log On


   



Question 616462: Please help me with this word problem: A person buys a cup of coffee and drinks it in the store. The coffee's temperature y(in degrees Fahrenheit) is given by y=70+137e^kt, where t is the number of minutes since he bought the coffee. A.) Use the equation to determine the temperature of the coffee when the person bought it. B.) If the person begins drinking the coffee when it reaches 180'f, use logarithms to find out how much time must he wait after buying it (start drinking it.)
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
A person buys a cup of coffee and drinks it in the store. The coffee's temperature y(in degrees Fahrenheit) is given by y=70+137e^kt, where t is the number of minutes since he bought the coffee.
.
A.) Use the equation to determine the temperature of the coffee when the person bought it.
When the person bought it, t is zero.
replace t with 0 and solve for y:
y=70+137e^kt
y=70+137e^k(0)
y=70+137e^0
y=70+137(1)
y=70+137
y=207 degrees Fahrenheit
.
B.) If the person begins drinking the coffee when it reaches 180'f, use logarithms to find out how much time must he wait after buying it (start drinking it.)
Here, they give you y as 180.
Replace y with 180 and solve for t:
y=70+137e^kt
180=70+137e^kt
110=137e^kt
110/137 = e^kt
ln(110/137) = kt
ln(110/137)/k minutes = t
.
The problem does not give enough information to define what the constant 'k' is