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

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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

Mathematics
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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.
Do you know what my x-value would be?
The correct answer is (1/3)ln(4) but I don't know how to get it

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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
I don't completely understand. Do you mind helping me solve it?
This is exponential growth\[A = Ae^{rt}\]
\[A = Pe^{rt}\] scratch the first one
Yeah same formula. Just put in the values you have and solve for the time

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