Question

Suppose you turn off the engine of your car when the temperature of the engine reaches 190 °F. If the outside temperature is a constant 65 °F, then the temperature T of the engine t minutes after you turn off the engine satisfies the equation below.
ln

T − 65
125

= −0.07t

(a) Solve the equation for T.

(b) Find the temperature of the engine 22 minutes after you turn it off. Give your answer to the nearest tenth of a degree.

Answer 1

\[\ln((T-65)/125)=-0.07t\]

Answer 2

first get rid of ln by e^(everything)

Answer 3

(T-65)/125=ln(-0.07t

Answer 4

or e^-0.07t

Answer 5

\[(T-65)/125=e^{-0.07t}\]

Answer 6

(T-65)/125 = e^(-0.07t) yea

then times 125 both sides and then add 65 both sides to get T alone

Answer 7

thank you?

Answer 8

o: you re welcome

you know how to do b, right?

Answer 9

so:
T = 125*e^(-0.07t) +65
(since t = 22 minutes)
T = 125*e^(-0.07*22) + 65
T = 125*e^(-1.54) + 65
T = 125*0.2143811 + 65
T = 91.79763 ºF

Answer 10

near to a tenth of a degree, T = 91.8 ºF