When performing water changes in a tank, each change removes a portion of the current water, not the original water. Thus, you cannot achieve exactly 100% with a finite number of changes. However, you can get arbitrarily close to 100%.
Let's explore the process step by step:
1. After the first 25% water change, 75% of the original water remains.
2. After the second 25% water change, 75% of the remaining 75% of the original water remains, which is \( 75\% \times 75\% = 56.25\% \).
3. After the third 25% water change, 75% of the remaining 56.25% of the original water remains, which is \( 75\% \times 56.25\% = 42.1875\% \).
This pattern can be described mathematically as:
\[ (0.75)^n \]
where \( n \) is the number of water changes.
To find the number of changes needed to reduce the original water to less than 1% of its original amount, solve for \( n \):
\[ 0.75^n < 0.01 \]
Take the logarithm of both sides:
\[ n \log(0.75) < \log(0.01) \]
\[ n > \frac{\log(0.01)}{\log(0.75)} \]
Using a calculator to find the values of the logarithms:
\[ \log(0.01) = -2 \]
\[ \log(0.75) \approx -0.125 \]
Therefore:
\[ n > \frac{-2}{-0.125} \]
\[ n > 16 \]
So, it takes more than 16 changes to get the original water reduced to less than 1% of its original amount.
For practical purposes, performing 17 changes will reduce the original water to less than 1%:
\[ 0.75^{17} \approx 0.0098 \]
So, you need 17 changes to reduce the original water content to less than 1%, which is practically very close to 100% removal.