help

Melissa:

how to solve for k?
150=80e^(-k)(0)+82

Hi Melissa,

I am not quit sure where you want me to put the 0 but no matter where I put it the question does not make sense.
One way I get 150=82 and the other way I get 150=162

I think you better check the question.

Sorry.... It should be this.

162=80e^-(k)(1) +82

k is the constant that I need to solve for. Room temp is 82. I have a thermos filled with liquid, original temp is 162. one hour later the temp of the liquid if 150. solve for K, rounding to the nearest 5th decimal place.

Oh my gosh, I messed it up again... I think.

I think I should make the formula : 150= 80e^-(k)(1) +82


Newton's law of heating and cooling. I'm so lost.. Thank you for the help!

Melissa:

Oh my gosh, I messed it up again... I think.

I think I should make the formula : 150= 80e^-(k)(1) +82


Newton's law of heating and cooling. I'm so lost.. Thank you for the help!

T(t) = A + B e ^-kt

T(1) = 82 + 80e ^-kt

150 = 82 + 80 e ^-k

Now I have got it to your starting point
surely you can at least start this Melissa
Can't you get e ^-k on one side by itself ?
After you do that then you will need to take the log (base e is best for this problem) of both sides. You always have to do this when the unknown is an exponent.
If you need to look up log identities, this is a good place
http://en.wikipedia.org/wiki/List_of_logarithmic_identities

Have a go and if you get stuck, post again.

T(t) = A + B e-kt

T(1) = 82 + 80e-kt

150 = 82 + 80 e-k

152-82 = 80 e^-k

70 = e^-k

ln 70 = ln e^-k

4.24849524205 = -k

k = -4.25