Bacterial Growth: How Long To Reach 6400 Bacteria?

Let's dive into a fascinating problem about bacterial growth! We're going to explore how to use a mathematical model to predict when a bacterial population will reach a specific size. This is a classic example of exponential growth, and understanding it can help us with various real-world applications, from medicine to environmental science. So, buckle up, guys, and let's get started!

Understanding the Problem

Our main goal is to determine the time it takes for a bacterial population to grow from an initial size of 100 to 6400, given the growth function F(t) = F₀ * (2/3)^t. It sounds a bit complicated, but don't worry, we'll break it down step by step. The function itself is crucial. It tells us how the number of bacteria, F(t), changes over time (t). Here, F₀ represents the initial number of bacteria, which is 100 in our case. The (2/3)^t part is the growth factor, indicating how the population changes with each passing time unit. It's important to realize this is a decay function, not a growth function, because the base (2/3) is less than 1. This means the bacterial population is actually decreasing over time, which is a twist! This initial observation is key because it means our final answer should reflect a time when the bacteria count is less than the initial 100, not more. The question as posed might contain an error, but we'll proceed with the mathematical steps to illustrate the process. Recognizing these details upfront is super important for preventing mistakes and ensuring our solution makes sense in the context of the problem.

We need to find the value of 't' (time in hours) when F(t) equals 6400. This involves setting up an equation and solving for 't'. Remember, the function F(t) = F₀ * (2/3)^t is the key to unlocking this problem. The initial number of bacteria, F₀, plays a vital role. In this scenario, F₀ = 100, meaning we started with a humble 100 bacteria. The (2/3) component dictates how the bacteria population evolves over time. Since this value is less than 1, it suggests a decay or reduction in the bacterial count rather than growth, which is a crucial point to keep in mind. The question states we want to find the time when there are 6400 bacteria. However, given that the function represents decay, we might encounter a situation where our mathematical solution doesn't align with the practical context. We should be prepared to interpret the results carefully and possibly identify any inconsistencies or errors in the problem statement itself. Solving mathematical problems isn't just about plugging in numbers; it's about critically thinking about the outcome and seeing if it makes logical sense.

Setting Up the Equation

Now, let's set up the equation to solve for 't': 6400 = 100 * (2/3)^t. This equation is the heart of our problem. It directly translates the problem statement into a mathematical form. On the left side, we have the target number of bacteria, 6400. On the right side, we have the initial number of bacteria (100) multiplied by the growth/decay factor (2/3)^t. Our mission is to isolate 't' and find its value. To do that, we will use the power of logarithms, but before we jump into logarithms, let's simplify the equation first. Dividing both sides by 100 makes the numbers easier to handle. This is a common strategy in problem-solving – simplifying before diving into complex operations. So, we divide both sides by 100, which gives us 64 = (2/3)^t. Now, this looks more manageable! We're one step closer to solving for 't'. Remember, the key to solving exponential equations like this is to use logarithms. Logarithms are the inverse operation of exponentiation, and they allow us to bring the exponent (which is 't' in our case) down from its lofty position. This is why understanding logarithms is essential for tackling these types of problems. We will apply logarithms in the next step to isolate 't' and find our solution.

Solving for t

To solve for 't', we'll first divide both sides by 100: 64 = (2/3)^t. Next, we'll take the logarithm of both sides. Using the natural logarithm (ln) is a common choice, but any logarithm base would work. Applying the natural logarithm gives us: ln(64) = ln((2/3)^t). Now, we use the logarithmic property that ln(a^b) = b * ln(a) to bring the exponent down: ln(64) = t * ln(2/3). This is a crucial step because it frees 't' from the exponent, making it possible to isolate. Now, to isolate 't', we divide both sides by ln(2/3): t = ln(64) / ln(2/3). This formula gives us the value of 't', which represents the time in hours. However, we still need to calculate the actual numerical value. Remember that ln(2/3) is negative because 2/3 is less than 1. This means that 't' will also be negative since ln(64) is positive. A negative value for 't' might seem strange in the context of time, but it aligns with our earlier observation that the bacterial population is decreasing. A negative time suggests we're looking at a point in the past relative to the initial observation. This is a really important check – it confirms that our mathematical steps are consistent with the nature of the problem (decay, not growth). We're almost there; we just need to crunch the numbers now!

Calculating the Time

Now, let's calculate the value of t: t = ln(64) / ln(2/3). Using a calculator, we find that ln(64) ≈ 4.1589 and ln(2/3) ≈ -0.4055. Therefore, t ≈ 4.1589 / -0.4055 ≈ -10.25 hours. So, according to our calculations, it would have taken approximately -10.25 hours to reach 6400 bacteria if the function accurately modeled growth from a smaller population to 6400. However, we know the function represents decay. This negative value indicates that 6400 bacteria was a population size in the past, relative to the starting point of 100 bacteria at t=0. This result is crucial! It highlights the importance of interpreting the mathematical solution within the context of the problem. We didn't just blindly calculate a number; we thought about what that number means. In this case, the negative time makes sense given the decaying nature of the bacterial population as modeled by the function. It also raises a very important point: The problem might be flawed in its initial setup. Asking when the population increases to 6400 when the model decreases the population leads to this logical inconsistency. Spotting these inconsistencies is a key skill in problem-solving. We’ve not only solved the mathematical equation but also critically evaluated the problem itself.

Interpreting the Result

Our final answer, t ≈ -10.25 hours, is a bit tricky because it's negative. As we've discussed, this means that if we were to extrapolate the decay backwards in time, the population would have been 6400 approximately 10.25 hours before Ulises started the experiment with 100 bacteria. This highlights a crucial point: the problem as stated seems to have a logical flaw, since the function describes a decreasing bacterial population, not an increasing one. The function F(t) = F₀ * (2/3)^t represents exponential decay, meaning the population decreases over time. If the initial population is 100, it's impossible for the population to increase to 6400 according to this model. The question is fundamentally asking about an event that contradicts the given mathematical model. This is a very common situation in real-world problem-solving. Sometimes, the data, assumptions, or even the problem statement itself can have errors or inconsistencies. Our job as problem-solvers isn't just to find a numerical answer but also to critically evaluate the entire situation and identify any potential issues. In this case, we've successfully used the math to identify a flaw in the problem itself. Guys, that's awesome problem-solving!

Conclusion

In conclusion, by using the given function and solving for 't', we found that the bacterial population would have been 6400 approximately 10.25 hours before the start of the experiment, if we were to extrapolate the model backwards. However, the most important takeaway here is that we identified a potential issue with the problem statement itself. The function describes a decaying population, making it impossible to reach a population of 6400 bacteria in the future from an initial population of 100. This exercise demonstrates not only how to solve exponential equations but also the importance of critical thinking and logical reasoning when tackling mathematical problems. Always remember, guys, math isn't just about numbers; it's about understanding the real-world context and making sure our solutions make sense!