Question

Suppose you turn off the engine of your car when the temperature of the engine reaches 191 °F. If the outside temperature is a constant 66 °F, then the temperature T of the engine t minutes after you turn off the engine satisfies the equation below.

ln

T − 66
125

= −0.07t

(a) Solve the equation for T.
T =


(b) Find the temperature of the engine 30 minutes after you turn it off. Give your answer to the nearest tenth of a degree.

°F

Answer 1

To solve the equation for T, we isolate T by rewriting the equation using the properties of logarithms. For part b, we substitute t = 30 into the equation to find the temperature of the engine 30 minutes after it is turned off.

Step-by-step explanation:

To solve the equation for T, we will isolate T by rewriting the equation using the properties of logarithms.

ln((T-66)/125) = -0.07t

(T-66)/125 = e^(-0.07t)

T-66 = 125 * e^(-0.07t)

T = 125 * e^(-0.07t) + 66

For part b, we substitute t = 30 into the equation to find the temperature of the engine 30 minutes after it is turned off.

T = 125 * e^(-0.07*30) + 66 = 244.2 °F

Answer 2

a)
Take Newton’s law of cooling as a basis and make a transformation :
From :
`ln(T-65/125) = -0.07t` to :
`ln[ (T - 65 ) / 125 ] = -0.07t`
Then solve :
`ln[ (T - 65 ) / 125 ] = -0.07t`

`(T - 65 ) / 125 = e^(-0.07t)`
And the result :
`T = 65 + 125e^(-0.07t)`

b) The easiest way how you can find the temperature of the engine 30 minutes after you turn it off :
`T = 65 + 125e^(-0.07t)`

`T = 65 + 125e^(-0.07*30)`
Here is the solution in the nearest tenth of a degree : T = 80.3℉