
assistance please
can any 1 help me get tru this single problem ...please
any help will be greatly accepted
A thermometer reading 11°C is brought into a room with a constant temperature of 20°C. If the thermometer
reads 18°C after 3 minutes, what will it read after being in the room for 5 minutes? Assume the cooling follows
Newton's Law of Cooling:
U = T + (U0  T)e^kt.

Use the curve given to you to find the value of the constants.
$\displaystyle U = T + (U_o  T)e^{kt}$
At time t = 3, the temperature is 18 C.
$\displaystyle 18 = 20 + (11  20)e^{3k}$
Find the value of k.
Then, use the equation again with the value of k you obtained and t = 5. FInd the value of U. (Smile)
What do you get?

$\displaystyle e^{kt}= left(e^k\right)^t$ so you really only need to find $\displaystyle e^k$, not k itself.

newton law of cooling
needs some assistance with this question please
A thermometer reading 11°C is brought into a room with a constant temperature of 20°C. If the thermometer
reads 18°C after 3 minutes, what will it read after being in the room for 5 minutes? Assume the cooling follows
Newton's Law of Cooling:
$\displaystyle
U = T + (U0  T)e^{kt}
$
(Round your answer to two decimal places.)
reach as far as this
$\displaystyle
18 = 20 + (11  20)e^{3k}
$
dont no if it is correct

I think you'll find that, with the way you've written the equation, k < 0. Try to solve for k. What do you get?


Ok. So far, so good. Now you need to "undo" the exponential function in order to get at the exponent. How do you "undo" an exponential?

not sure how thats done
can u break it down futher5 for me please

What is the function inverse of the exponential function? If I have the equation $\displaystyle y=e^{x},$ how could I solve for $\displaystyle x?$

is it this
$\displaystyle
ln (\frac{2}{9}) = ln (e)^{3k}
$
$\displaystyle
3k ln (e) = ln (\frac{2}{9})
$
$\displaystyle
3k ln (e) = ln (\frac{2}{9})
$
$\displaystyle
3k = 1.5040
$
$\displaystyle
k = 0.501359133
$

Well, I would write it this way (being careful with parentheses):
$\displaystyle \ln\left(\dfrac{2}{9}\right)=\ln\left(e^{3k}\right ).$
But that's definitely the right idea. Now what?

$\displaystyle
1.5040774 = 3kln (e)
$


$\displaystyle
1.5040774 = 3k
k = 0.501359133
$

So, k = 0.501359133. Now what?