The number of cells of a certain type of bacteria increases continuously at a rate equal to two
more than three times the number of bacteria present. If there are 10 present initially, and 42 present T hours
later, find the exact value of T

Question

The number of cells of a certain type of bacteria increases continuously at a rate equal to two 
more than three times the number of bacteria present.  If there are 10 present initially, and 42 present T hours 
later, find the exact value of T

anonymous · Answer

It's a simple linear equation, relating y as a function of t.

For example, you have y as the number of bacteria, and t as hours.

Therefore, using the formula y=mx + b, you get y=(2+ 3y)t + 10

Solve for t by substituting in 42 for y and trucating.

anonymous · Answer

Do you know what my x-value would be?

anonymous · Answer

The correct answer is (1/3)ln(4) but I don't know how to get it

anonymous · Answer

You don't have an x value. I used y in my example because It is the output, but the independent variable is t.

If their answer is in terms of natural log then you must go about it more specifically, using the formula k=Ce^rt

anonymous · Answer

I don't completely understand. Do you mind helping me solve it?

anonymous · Answer

This is exponential growth$$A = Ae^{rt}$$

anonymous · Answer

$$A = Pe^{rt}$$ scratch the first one

anonymous · Answer

Yeah same formula. Just put in the values you have and solve for the time