The number of fish in a lake decreases by x% each year.
Given that the number of fish halves in 8 years, work out the value of x.
Give your answer correct to 1 decimal place.

To solve for the value of \( x \), we can model the problem using exponential decay. The number of fish decreases by \( x \%\) each year, so after 1 year, the number of fish is multiplied by \( (1 - \frac{x}{100}) \).

Given that the number of fish halves in 8 years, we can set up the equation:

\[
N \times \left(1 - \frac{x}{100}\right)^8 = \frac{N}{2}
\]

Here, \( N \) is the initial number of fish. Simplifying the equation:

\[
\left(1 - \frac{x}{100}\right)^8 = \frac{1}{2}
\]

To solve for \( x \), follow these steps:

1. Take the natural logarithm (ln) of both sides:

\[
\ln\left(\left(1 - \frac{x}{100}\right)^8\right) = \ln\left(\frac{1}{2}\right)
\]

2. Apply the logarithm power rule:

\[
8 \cdot \ln\left(1 - \frac{x}{100}\right) = \ln\left(\frac{1}{2}\right)
\]

3. Divide both sides by 8:

\[
\ln\left(1 - \frac{x}{100}\right) = \frac{\ln\left(\frac{1}{2}\right)}{8}
\]

4. Use the logarithm value \( \ln\left(\frac{1}{2}\right) = -\ln(2) \approx -0.6931 \):

\[
\ln\left(1 - \frac{x}{100}\right) = \frac{-0.6931}{8} \approx -0.08664
\]

5. Solve for \( 1 - \frac{x}{100} \):

\[
1 - \frac{x}{100} = e^{-0.08664} \approx 0.916
\]

6. Finally, solve for \( x \):

\[
\frac{x}{100} = 1 - 0.916 = 0.084
\]

\[
x = 0.084 \times 100 = 8.4
\]

Thus, the value of \( x \) is approximately \( 8.4\%\) to one decimal place.

---

Do you want more details on any step? Here are 5 related questions you might consider:

1. How would the value of \( x \) change if the fish population halved in 10 years instead of 8 years?
2. What would be the number of fish after 5 years, given this rate of decline?
3. Can you express the decay rate as a continuous percentage rate instead of a yearly percentage rate?
4. How would you model the situation if the fish population increased by a percentage each year instead?
5. What happens to the population if the decrease is not by a constant percentage but by a fixed number each year?

**Tip:** When dealing with exponential decay or growth, always consider how logarithms can simplify equations involving powers.

Learn how to calculate the annual decline rate of a fish population using exponential decay. Solve for the percentage decrease per year when the fish population halves in 8 years. Step-by-step solution with logarithmic calculations provided.

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