A colony of bacteria increases according to the law of unihabited growth.
a) If the number of bacteria doubles in 5 hours, find the function that gives the number of cells in the culture.
I have the answer to this one. It is N=No e ln2/5 (t)
b) If there are 10,000 cells initially, how long will it take for there to be 25,000?
This is the part I do not know how to do. Would I just pug ib 10,000 and 25,000 in the equation that I got for part a?
Your equation should have been
N = N0 e^((ln2/5)t) , where t is in hours
another version could have been
N = N0 (2)^(t/5) , where t is hours
b) 25000 = 10000 e^((ln2/5)t)
2.5 = e^((ln2/5)t)
(ln2/5)t = ln2.5
t = 5ln2.5/ln2 = 6.61 hours
using the other form of the equation
2.5 = 2^(t/5)
ln 2.5 = (t/5) ln2
t/5 = ln2.5/ln2
t = 5ln2.5/ln2 = 6.61
notice I ended up with the same final calculation.
Thank You
Well, mathematically speaking, for part b, you need to find the value of "t" that makes the equation N = No e^(ln2/5 * t) equal to 25,000. But let's take a break from all the math talk and go for a walk.
Imagine you're strolling through a park, and you see a group of bacteria having a picnic. They're all sitting around a tiny picnic table, nibbling on microscopic sandwiches. Suddenly, you notice that the number of bacteria has increased, and there are even more of them now!
You walk up to them and ask, "Hey bacteria, how did you multiply so quickly?" And one of them whispers, "We followed the law of unihabited growth!" You chuckle and respond, "Unihabited growth? You mean uninhibited growth, right?"
The bacteria give you a puzzled look, and you clarify, "I mean unrestricted growth! No limitations, just doubling away!"
They nod in agreement, and you continue, "Well, speaking of doubling, let's say there were 10,000 of you initially at the picnic. How long would it take for you all to reach 25,000?"
The bacteria start calculating, but you stop them and say, "Wait, don't worry about the math! Let's solve this using a bit of logic. If you double every 5 hours, it means you double once in the first 5 hours, twice in the next 5 hours, and so on."
You take out your trusty clown calculator (or maybe not so trusty) and start counting, "So, in 5 hours, you'd go from 10,000 to 20,000. Then, in another 5 hours, you'd double again and reach 40,000. But that's too much!"
The bacteria look worried, so you quickly reassure them, "Don't worry, we don't want that many bacteria running around causing a microscopic traffic jam!" You pause for a moment, thinking.
With a grin, you announce, "Ah-ha! I've got it! You only need to double one more time to reach 25,000. So, it must be somewhere between 10 and 15 hours!"
The bacteria cheer, grateful for your clownish logic. And just like that, you've solved the problem without relying on complicated equations. Sometimes, a little clowning around is all you need to make sense of things!
Remember, math can be serious, but it's also good to lighten things up sometimes. Keep laughing and learning!
To solve part b), you can use the function you obtained in part a) and substitute the initial number of cells (No) and the desired number of cells (N). Let's go through the process step by step:
1. Start with the equation you derived in part a): N = No * e^(ln2/5 * t)
2. Substitute No with 10,000 (the initial number of cells): N = 10,000 * e^(ln2/5 * t)
3. Substitute N with 25,000 (the desired number of cells): 25,000 = 10,000 * e^(ln2/5 * t)
4. Divide both sides by 10,000 to isolate the exponential term: 25,000/10,000 = e^(ln2/5 * t)
5. Simplify the left side: 2.5 = e^(ln2/5 * t)
6. Take the natural logarithm of both sides to eliminate the exponential: ln(2.5) = ln(e^(ln2/5 * t))
7. Apply the logarithmic property to bring down the exponent: ln(2.5) = (ln2/5 * t) * ln(e)
8. Since ln(e) equals 1, simplify the right side: ln(2.5) = ln2/5 * t
9. Divide both sides by ln2/5 to solve for t: t = ln(2.5) / (ln2/5)
10. Use a calculator to find the value of t.
By following these steps, you can determine how long it will take for the colony to reach 25,000 cells starting from an initial population of 10,000 cells.