To find the number of hours it will take for the dye to reach 10% of its original brightness, we can first express the brightness after \( n \) hours using the given recursive formula, which initializes at \( B(0) = 100 \).
Using the recursive relationship:
\[ B(n) = B(n-1) \cdot 0.80 \]
we can derive the explicit expression for \( B(n) \):
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Calculate \( B(1) \): \[ B(1) = B(0) \cdot 0.80 = 100 \cdot 0.80 = 80 \]
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Calculate \( B(2) \): \[ B(2) = B(1) \cdot 0.80 = 80 \cdot 0.80 = 64 \]
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Calculate \( B(3) \): \[ B(3) = B(2) \cdot 0.80 = 64 \cdot 0.80 = 51.2 \]
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Continue this pattern until we find when \( B(n) \leq 10 \).
For generalization, we can express \( B(n) \) in terms of \( n \):
\[ B(n) = 100 \cdot (0.80)^n \]
Next, we need to solve for \( n \) when \( B(n) \) reaches 10% of the original brightness:
\[ 100 \cdot (0.80)^n = 10 \]
Dividing both sides by 100:
\[ (0.80)^n = 0.1 \]
Now we can take the logarithm of both sides to solve for \( n \):
\[ \log((0.80)^n) = \log(0.1) \]
Using the property of logarithms that allows us to bring the exponent down, we have:
\[ n \cdot \log(0.80) = \log(0.1) \]
Solving for \( n \):
\[ n = \frac{\log(0.1)}{\log(0.80)} \]
Now, we will calculate the values:
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Calculate \( \log(0.1) \): \[ \log(0.1) = -1 \quad (\text{base 10 logarithm, since } 10^{-1} = 0.1) \]
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Calculate \( \log(0.80) \): Using a calculator, we find: \[ \log(0.80) \approx -0.09691 \]
Now substituting back in:
\[ n = \frac{-1}{-0.09691} \approx 10.33 \]
Since \( n \) must be a whole number (as it represents hours), we round up to the next whole hour. Therefore, it will take approximately 11 hours for the dye to reach 10% of its original brightness.
Thus, the answer is:
\[ \boxed{11} \]