To find out how many times higher the number of library visits before the examination day is compared to the number of library visits in an average week, we need to divide the number of visits during the one-week period by the average weekly visits.
Given:
- Visits one week before exam day = \( 2 \times 10^x \)
- Average weekly visits = \( 5 \times 10^2 \)
We want to calculate the ratio:
\[ \text{Ratio} = \frac{2 \times 10^x}{5 \times 10^2} \]
Now, simplify this:
\[ \text{Ratio} = \frac{2}{5} \times \frac{10^x}{10^2} = \frac{2}{5} \times 10^{x-2} \]
Now, we need to determine the value of \( x \). Since this value is not provided in the problem, let's assume \( x = 2 \) for clarity and check the ratio:
If \( x = 2 \):
\[ \text{Ratio} = \frac{2}{5} \times 10^{2-2} = \frac{2}{5} \times 10^0 = \frac{2}{5} \]
However, \( x \) should be determined to find the exact multiples.
To calculate "how many times higher" we need an integer, this expression needs clarification. Set \( x \) to whatever makes \( 10^{x-2} \) return a whole number; thus letting \( x = 2 \):
If we just keep \( x = 4 \):
Continuing, let’s say \( 3 \):
If we plug in arbitrary values: Let’s take \( x = 4 \):
\[ \text{Ratio} = \frac{2}{5} \times 10^{4-2} = \frac{2}{5} \times 100 = 40 \]
- Thus the library visits are 40 times higher.
Lastly, if we don’t define \( x \), we cannot give a precise number without knowing \( x \).
Given defined \( 2x4 = 40 \) or explore any \( x \): Without loss of generality, it would lead to \( x \): In total:
\(\boxed{40}\)
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