A cup of coffee with the temperature of 90°C is set to cool in a room with a temperature of 20°C. T is the temperature of the coffee as a function of time in minutes. Find the temperature after 6 minutes if the coffee's cooling speed is 5% of the difference between the mug and the room.

I expressed it as a differential equation T'=-0.05(T-20). When I solved it, I got that T=-310/e^(0.05x)+400. If x=6, I'm getting that the temperature is greater than 90°. What am I doing wrong?
I'm not sure where the -310 or +400 came from, but somewhere in your solving of it something went wrong, because those should be +70 and +20 respectively.  My guess is that something went wrong with the +C part?  Remember that when you have a ln on one side of the equation and you want to turn it into an exponential, the *entire* side of the equation goes into the exponent, which will include the +C part.
Define F=T-20, then F'=T'-20'=T'-0=T'. Therefore F'=-F/20, so F=C\*exp(-x/20). Substituting T back gives you T=C\*exp(-x/20)+20. You can solve C from the initial condition.
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