Here are my steps: u(t)=t+(uo-t)e^kt where k< 0 300=70+(450-70)e^5k 300=70+(380)e^5k 230=380e^5k e^5k=23/38 5k lne= 23/38 k=ln 23/38/5 k=-.1004183888 230=70+380e^5k 380e^5k=160 e^kt=8/19 kt=ln 8/19 kt/k=ln 8/19/k t=8.6 minutes. 8.6 minutes + 5 minutes initial cooling time = 5: 13 pm.

You have the correct approximate value of k. The part after that is where things get a bit strange.

You don't plug in t = 5 again. We don't know the value of t yet. We're trying to solve for it.

We do know u(t) is 135 as we want the temperature to be 135 degrees. Plug this in and isolate t. 

So 
u(t) = 70+380e^(-0.1004183888t)
135 = 70+380e^(-0.1004183888t)
135-70 = 380e^(-0.1004183888t)
65 = 380e^(-0.1004183888t)
380e^(-0.1004183888t) = 65
e^(-0.1004183888t) = 65/380
e^(-0.1004183888t) = 13/76
-0.1004183888t = ln(13/76)
-0.1004183888t = -1.7657839828248
t = -1.7657839828248/(-0.1004183888)
t = 17.5842692152893

So it takes about 17.58 minutes for the item to cool to the desired temp

The answer is d, none of above hope i