Calc question using e?

Coffee question

T = 80+85e^-0.045t
T= degrees farenheit
t= time (minutes)

a. what was the temperature of the coffee when it was poured

b. When will the coffee be 120 degrees farenheit

cidyah2016-10-19T06:18:09Z

a)
substitute t=0
T = 80+85 e^0
T = 80+85 = 165 °F

b)
solve:
120 = 80 + 85 e^(-0.045 t)
120-80 = 85 e^(-0.045t)
85 e^(-0.045t) = 40
e^(-0.045t) = 40/85
-0.045 t = ln (40/85)
t = -ln(40/85) / 0.045
t = 16.75 (time at which coffee will be 120 °F

Brainard2016-10-19T05:59:22Z

a. what was the temperature of the coffee when it was poured

At t = 0

T = 80 + 85e^0

= 80 + 85

= 165 ° F


b. When will the coffee be 120 degrees farenheit

When T = 120

120 = 80 + 85e^(-0.045t)

40 = 85e^(-0.045t)

40e^(0.045t) = 85

e^(0.045t) = 85/40

= 2.125

Taking logs to base e

0.045t = ln(2.125)

t = ln(2.125)/0.045

= 16.7504845

= 16.75 minutes