The given problem states that the intensity $I$ of light passing through water is modeled by the equation:

$I = 10^{1 - 0.13x}$

where $x$ is the depth of the water in meters. It also mentions that most aquatic plants require a light intensity of $4.2$ units for strong growth. We need to determine the depth $x$ at which this intensity is met.

Solution

1. Set the equation equal to the required intensity: $4.2 = 10^{1 - 0.13x}$

2. Take the base-10 logarithm of both sides to solve for $x$: $\log_{10}(4.2) = \log_{10}(10^{1 - 0.13x})$

   Using the logarithmic property $\log_{10}(10^a) = a$, we get: $\log_{10}(4.2) = 1 - 0.13x$

3. Calculate $\log_{10}(4.2)$: From a calculator: $\log_{10}(4.2) \approx 0.6232$

   Substituting this back into the equation: $0.6232 = 1 - 0.13x$

4. Rearrange to isolate $x$: $0.13x = 1 - 0.6232$

   Simplify the right-hand side: $0.13x = 0.3768$

   Divide through by 0.13: $x = \frac{0.3768}{0.13}$

5. Final calculation: $x \approx 2.9 \, \text{meters}$

Conclusion

The depth of water at which the light intensity is 4.2 units is approximately 2.9 meters.

Would you like further details or explanations about any step?

Here are 5 related questions to deepen understanding:

1. How can the logarithmic properties be applied to solve exponential equations in general?
2. What happens to the intensity as the depth increases?
3. How does the constant $0.13$ in the exponent affect the model?
4. What would the depth be if the required intensity were 2 units instead?
5. How can this equation be used to study light penetration in other liquids?

Tip: When solving equations involving logarithms, always double-check your base and ensure proper handling of decimal approximations for accuracy!