Question

A bowl of soup cools according to Newtons law of cooling. Its temperature (in degrees farenheit) at time t is given by T(t)=68+144e^0.04t. Where t is given in minutes.
How long after serving is the soup 125 degrees?

Answer 1

it should be \[\Large T(t)=68+144e^{-0.04t}\]
the exponent is -0.04t and not 0.04t

Answer 2

Replace T(t) with 125 and solve for t
\[\Large T(t)=68+144e^{-0.04t}\]
\[\Large 125=68+144e^{-0.04t}\]
\[\Large 125-68=144e^{-0.04t}\]
\[\Large 57=144e^{-0.04t}\]
\[\Large \frac{57}{144} = e^{-0.04t}\]
\[\Large 0.3958333 = e^{-0.04t}\]
I'll let you finish up

Answer 3

Thank you.

Answer 4

you're welcome

Answer 5

wait so I get .395/-0.04 which is -9 something, but you cant have negatives as minutes that have passed

Answer 6

@jim_thompson5910

Answer 7

how do you undo exponents? which type of function do you need to apply?

Answer 8

I tried rounding uo to .4/-0.04 but I just got -10

Answer 9

Oh yeah you have to use logs

Answer 10

specifically, log with base e aka LN

\[\Large \log_{e} = \ln\]

Answer 11

So then ln(.395)/-0.04?

Answer 12

correct

Answer 13

what is that equal to?

Answer 14

23.22 minutes:)

Answer 15

very good

Answer 16

sweet.