I need help. I don't even know where to start

[Jessebeth]

I hope someone can help me with this equation. I have seriously been working on my homework for 10 hours now. I do not know where to start on this question. I keep coming up with T=20+200e^(-.11t) for part A. If thats right, then I have no idea where to go with part b. Someone please help me. Thank you so much.

Suppose you’re driving your car on a winter day (20 deg F outside) and the
engine overheats (at about 220 deg F). When you park, the engine begins to cool
down. The temperature T, of the engine, t minutes after you park, satisfies the
equation:

ln(T-20/200)=-.11t

a) Solve this equation for T.
b) Use part a) to find the temperature of the engine after 20 min. (t=20).

 

[tkhunny]

I do not know where to start on this question.
I have no idea where to go with part b.

Hogwash. You're nearly done. Have you considered simply substituting t = 20 minutes into your equation?

 

[mmm4444bot]

T = 20 + 200e^(-.11t)

This expression for T is correct, but you forgot to type grouping symbols around the numerator in the given equation. I've added them in red below.


ln((T-20)/200)=-.11t


b) Use part a) to find the temperature of the engine after 20 min. (t=20).

Part (b) requires you to calculate the value of T when t = 20.

T = 20 + 200 e^(-0.11*20)