Question

try a few examples to verify that the doubling time is indeed 7 days.
day 0: 4 ✔️ plants
day 7: 8 ✔️ plants
did the water lilies double? ✔️ yes
day 10: 10 ✔️ plants
day 17: 22 ✔️ plants
did the water lilies double? ✔️ approximately
complete
the slider only goes up to day 30, but this pond can hold up to 400 water lilies. estimate the day that the pond will be full. day
done
day:
total water lilies: 4

Explanation:

Step1: Identify the growth pattern

The water lilies double every 7 days (from Day 0: 4, Day 7: 8, which is \(4\times2 = 8\)). So the growth is exponential with a doubling time of 7 days. The formula for exponential growth is \(N(t)=N_0\times2^{\frac{t}{T}}\), where \(N_0 = 4\) (initial number), \(T = 7\) (doubling time), and \(N(t)\) is the number at time \(t\).

Step2: Find \(t\) when \(N(t)=400\)

We set up the equation \(400 = 4\times2^{\frac{t}{7}}\). First, divide both sides by 4: \(\frac{400}{4}=2^{\frac{t}{7}}\), so \(100 = 2^{\frac{t}{7}}\).

Step3: Solve for \(t\)

Take the logarithm of both sides. Using base - 2 logarithm: \(\log_2(100)=\frac{t}{7}\). We know that \(\log_2(64) = 6\) (since \(2^6 = 64\)) and \(\log_2(128)=7\) (since \(2^7 = 128\)). Since \(100\) is between \(64\) and \(128\), \(\log_2(100)\approx6.64\). Then \(t = 7\times\log_2(100)\approx7\times6.64 = 46.48\). But we can also use a trial - and - error approach with the doubling time.

Let's list the number of water lilies at each doubling period:
- Day 0: 4
- Day 7: \(4\times2 = 8\)
- Day 14: \(8\times2 = 16\)
- Day 21: \(16\times2 = 32\)
- Day 28: \(32\times2 = 64\)
- Day 35: \(64\times2 = 128\)
- Day 42: \(128\times2 = 256\)
- Day 49: \(256\times2 = 512\)

Since at Day 42, we have 256 water lilies and at Day 49 we have 512 (which is more than 400), we can do a more precise calculation. The difference between 400 and 256 is \(400 - 256 = 144\). The growth from Day 42 to Day 49 is \(512 - 256 = 256\) over 7 days. The fraction of the 7 - day period needed is \(\frac{144}{256}=\frac{9}{16}\approx0.5625\). So the additional days needed is \(7\times0.5625\approx4\). So total days \(t\approx42 + 4=46\). But looking at the pattern from the given days (Day 0:4, Day 7:8, Day 14:16, Day 21:32, Day 28:64, Day 35:128, Day 42:256, Day 49:512). Since 256 (Day 42) and 512 (Day 49), and 400 is closer to 256. But let's check the initial doubling time. Wait, the problem says "the slider only goes up to day 30, but this pond can hold up to 400 water lilies. Estimate the day that the pond will be full." Wait, maybe we made a mistake. Wait, the initial number is 4. Let's use the formula \(N(t)=4\times2^{\frac{t}{7}}\). We need to find \(t\) when \(N(t) = 400\).

\(4\times2^{\frac{t}{7}}=400\)

\(2^{\frac{t}{7}} = 100\)

Taking natural logarithm: \(\ln(2^{\frac{t}{7}})=\ln(100)\)

\(\frac{t}{7}\times\ln(2)=\ln(100)\)

\(t=\frac{7\times\ln(100)}{\ln(2)}\)

\(\ln(100)\approx4.605\), \(\ln(2)\approx0.693\)

\(t=\frac{7\times4.605}{0.693}=\frac{32.235}{0.693}\approx46.5\)

But let's check the given data points: Day 0:4, Day 7:8, Day 14:16, Day 21:32, Day 28:64, Day 35:128, Day 42:256, Day 49:512. Since 256 is at Day 42 and 512 at Day 49, and 400 is between them. But maybe the problem expects us to use the doubling time and see that from Day 0 (4) to Day \(t\), we need to reach 400. Let's see how many times we double:

\(4\times2^n=400\)

\(2^n = 100\)

\(n=\log_2(100)\approx6.64\)

Since each \(n\) corresponds to 7 days, \(t = 7\times6.64\approx46.5\), so approximately Day 47. But maybe there is a simpler way. Wait, the problem's initial data: Day 0:4, Day 7:8, Day 14:16, Day 21:32, Day 28:64, Day 35:128, Day 42:256, Day 49:512. Since 256 (Day 42) and 512 (Day 49), and 400 is 400 - 256 = 144 more than 256. The rate of growth from Day 42 to Day 49 is 256 over 7 days, so per day growth is \(256\div7\approx36.57\). To get 144, we need \(144\div36.57\approx3.94\) days, so total days is \(42 + 4 = 46\), so approximately Day 46 or 47. But let's check the problem again. Wait, maybe the doubling time is 7 days, so let's list the number of water lilies:

- Day 0: 4
- Day 7: 8 (4*2)
- Day 14: 16 (8*2)
- Day 21: 32 (16*2)
- Day 28: 64 (32*2)
- Day 35: 128 (64*2)
- Day 42: 256 (128*2)
- Day 49: 512 (256*2)

Since 512 is more than 400 and 256 is less than 400, and 400 is closer to 512? No, 400 - 256 = 144, 512 - 400 = 112. So it's closer to Day 49? Wait, no, 144>112? No, 144 is the difference from 256, 112 is the difference from 512. So 400 is closer to 512. So \(t\) is between 42 and 49. Let's use linear approximation between Day 42 (256) and Day 49 (512). The function \(N(t)\) is exponential, but for approximation, we can use linear. The slope \(m=\frac{512 - 256}{49 - 42}=\frac{256}{7}\approx36.57\). We need to find \(\Delta t\) such that \(256+m\times\Delta t = 400\). So \(\Delta t=\frac{400 - 256}{m}=\frac{144}{\frac{256}{7}}=\frac{144\times7}{256}=\frac{1008}{256}=\frac{63}{16}=3.9375\approx4\). So \(t = 42 + 4 = 46\). So approximately Day 46 or 47. But maybe the answer is around Day 46 - 47. But let's check the formula again. \(N(t)=4\times2^{\frac{t}{7}}\). Let's plug t = 46: \(4\times2^{\frac{46}{7}}=4\times2^{6.571}\). \(2^{6}=64\), \(2^{0.571}\approx2^{0.5 + 0.071}=\sqrt{2}\times2^{0.071}\approx1.414\times1.051\approx1.486\). So \(2^{6.571}\approx64\times1.486\approx95.1\), \(4\times95.1 = 380.4\). t = 47: \(4\times2^{\frac{47}{7}}=4\times2^{6.714}\). \(2^{0.714}\approx2^{0.7+0.014}\). \(2^{0.7}\approx1.624\), \(2^{0.014}\approx1.01\), so \(2^{0.714}\approx1.624\times1.01\approx1.64\). \(2^{6.714}\approx64\times1.64\approx104.96\), \(4\times104.96 = 419.84\approx420\). So at t = 47, we are close to 400 - 420. So the estimated day is around 47.

Answer:

Approximately Day 47 (or a value around 46 - 47, depending on the approximation method)