To solve this problem, we will use the following logic:

1. Understand the penny doubling pattern: The number of pennies doubles each day. The total number of pennies on day $n$ is given by: $P_n = 2^{n-1}$

2. Calculate the height of the stack: The height of the stack of pennies on day $n$ is: $H_n = P_n \times 1.5 \, \text{mm}$

   Substitute $P_n$: $H_n = 2^{n-1} \times 1.5 \, \text{mm}$

3. Convert the target distance to millimeters: The target distance is $3 \times 10^{13} \, \text{km}$. Converting to millimeters: $\text{Target height} = 3 \times 10^{13} \, \text{km} \times 10^6 \, \text{mm/km} = 3 \times 10^{19} \, \text{mm}$

4. Set up the equation to solve for $n$: $2^{n-1} \times 1.5 = 3 \times 10^{19}$

   Rearranging for $n$: $2^{n-1} = \frac{3 \times 10^{19}}{1.5}$

   Simplify: $2^{n-1} = 2 \times 10^{19}$

5. Take the logarithm to solve for $n$: Taking the base-2 logarithm of both sides: $n-1 = \log_2(2 \times 10^{19})$

   Use logarithmic rules: $n-1 = \log_2(2) + \log_2(10^{19})$

   Since $\log_2(2) = 1$ and $\log_2(10^{19}) = 19 \cdot \log_2(10)$, where $\log_2(10) \approx 3.32193$: $n-1 = 1 + 19 \cdot 3.32193$

   Calculate: $n-1 \approx 1 + 63.11667 = 64.11667$

   Add 1 to find $n$: $n \approx 65.12$

Final Answer:

It would take approximately 65 days for the stack of pennies to reach the star.

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Do you want more details on any part of this solution? Here are 5 related questions to expand your understanding:

1. How does the height of the penny stack change with exponential growth?
2. Can this doubling pattern be visualized using graphs?
3. What happens if the star were further away or closer?
4. How does logarithmic calculation help in solving real-world exponential growth problems?
5. What are other examples of problems involving doubling sequences?

Tip: Exponential growth is surprisingly fast; understanding it is key to grasping phenomena like compound interest and population growth