Question
A certain type of fluorescent dye used in scientific research loses its brightness over time when exposed to light. Initially, the brightness is at 100% of its original intensity. Each hour, the brightness decreases to 80% of the intensity of the previous hour. The recursive formula describing this decrease is B(n)=B(nā1)ā
0.80
, where B(n)
is the brightness as a percentage of the original intensity after š hours, and B(0)=100
. Determine the number of hours it will take for the dye to reach 10% of its original brightness.
, where B(n)
is the brightness as a percentage of the original intensity after š hours, and B(0)=100
. Determine the number of hours it will take for the dye to reach 10% of its original brightness.
Answers
Answered by
GPT-4o mini
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 \( B(n) \):
1. Calculate \( B(1) \):
\[
B(1) = B(0) \cdot 0.80 = 100 \cdot 0.80 = 80
\]
2. Calculate \( B(2) \):
\[
B(2) = B(1) \cdot 0.80 = 80 \cdot 0.80 = 64
\]
3. Calculate \( B(3) \):
\[
B(3) = B(2) \cdot 0.80 = 64 \cdot 0.80 = 51.2
\]
4. 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:
1. Calculate \( \log(0.1) \):
\[
\log(0.1) = -1 \quad (\text{base 10 logarithm, since } 10^{-1} = 0.1)
\]
2. 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:
\[
\boxed{11}
\]
Using the recursive relationship:
\[
B(n) = B(n-1) \cdot 0.80
\]
we can derive the explicit expression for \( B(n) \):
1. Calculate \( B(1) \):
\[
B(1) = B(0) \cdot 0.80 = 100 \cdot 0.80 = 80
\]
2. Calculate \( B(2) \):
\[
B(2) = B(1) \cdot 0.80 = 80 \cdot 0.80 = 64
\]
3. Calculate \( B(3) \):
\[
B(3) = B(2) \cdot 0.80 = 64 \cdot 0.80 = 51.2
\]
4. 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:
1. Calculate \( \log(0.1) \):
\[
\log(0.1) = -1 \quad (\text{base 10 logarithm, since } 10^{-1} = 0.1)
\]
2. 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:
\[
\boxed{11}
\]