Estimate the number of minutes that it will take the coffee temperature to reach 100 degrees if the degrees continues to drop 4% each minute

After 1 minute the temperature will be 0.96*150

After 2 minutes the temperature will be 0.96²*150

After 3 minutes the temperature will be 0.96³*150

...

After t minutes the temperature will be 0.96^t*150

What is t when the temperature is 100?

0.96^t*150 = 100

Take logs of both sides

log(0.96^t)+log(150) = log(100)

t*log(0.96) = log(100)-log(150) = log(100/150) = log(2/3)

t = log(2/3)/log(0.96)

$${\mathtt{t}} = {\frac{{log}_{10}\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{{log}_{10}\left({\mathtt{0.96}}\right)}} \Rightarrow {\mathtt{t}} = {\mathtt{9.932\: \!515\: \!862\: \!422\: \!638\: \!3}}$$

t ≈ 9.93 minutes

I think we need to know what temperature the coffee is at presently.

I forgot to put that in there.  The starting temp is 150 degrees

This is a little beyond me. I have an idea how to solve it but I'd rather you have someone who is familiar with questions like these to solve this for you. :)