To solve for $m$ such that the noise level $S(d) = 20$ decibels at a distance $m$ meters from the turbine, given the noise function $S(d) = a \cdot b^d$, we first need to determine the values of $a$ and $b$ using the information provided:

1. At $d = 0$, $S(0) = 105$ decibels.
2. At $d = 100$, $S(100) = 49$ decibels.

Step 1: Set up equations based on known values

From the first condition: $S(0) = a \cdot b^0 = a = 105$

This gives us $a = 105$.

Using the second condition, we can substitute $a = 105$ into the equation for $S(100)$: $S(100) = 105 \cdot b^{100} = 49$

Step 2: Solve for $b$

Divide both sides by 105: $b^{100} = \frac{49}{105}$

Simplify the fraction: $b^{100} = \frac{7}{15}$

Now, take the 100th root of both sides to solve for $b$: $b = \left( \frac{7}{15} \right)^{\frac{1}{100}}$

Step 3: Solve for $m$ when $S(m) = 20$

Now we want to find $m$ such that $S(m) = 20$. Substitute $a = 105$ and the expression for $b$: $20 = 105 \cdot \left( \frac{7}{15} \right)^{\frac{m}{100}}$

Divide both sides by 105: $\frac{20}{105} = \left( \frac{7}{15} \right)^{\frac{m}{100}}$

Simplify the left side: $\frac{4}{21} = \left( \frac{7}{15} \right)^{\frac{m}{100}}$

Step 4: Take the logarithm of both sides to solve for $m$

Using natural logarithms: $\ln \left( \frac{4}{21} \right) = \frac{m}{100} \cdot \ln \left( \frac{7}{15} \right)$

Solve for $m$: $m = 100 \cdot \frac{\ln \left( \frac{4}{21} \right)}{\ln \left( \frac{7}{15} \right)}$

Let’s calculate this value to find $m$.The distance $m$ at which the noise level is 20 decibels is approximately 217.58 meters from the turbine.

Would you like further details on this calculation or have any other questions?

Here are some related questions:

1. How does the noise level decrease as distance increases?
2. What would the noise level be at 300 meters?
3. How would the function change if the turbine’s noise at 0 meters increased?
4. How sensitive is the noise level to changes in the parameter $b$?
5. Can this model be used for other types of sounds or machinery?

Tip: When working with exponential models, logarithms are often helpful for solving equations where the variable is in the exponent.