## Comparing two numbers written in index form like $2^{11}$ and $3^7$ is not difficult, as any calculator would confirm that $2^{11} = 2048 \lt 3^7 = 2187$.

However, confirming that $632382^{518061} \gt 519432^{525806}$ would be much more difficult, as both numbers contain over three million digits.

Using base_exp.txt (right click and ‘Save Link/Target As…’), a 22K text file containing one thousand lines with a base/exponent pair on each line, determine which line number has the greatest numerical value.

NOTE: The first two lines in the file represent the numbers in the example given above.

### The entire problem statement seems to model a specific problem from Project Euler (#99).

The key to solve this problem involves understanding properties of logarithms – primarily that they allow us to convert operations of multiplication and exponentiation to addition and multiplication respectively.

Given two numbers in the form of base^exponent, `a^b` and `c^d`, and we want to compare which one is bigger without actually calculating the value (which is quite resource-intensive for large numbers).

Using properties of logarithms, we can transform these numbers into a more manageable form.

Specifically, given our numbers are `a^b` and `c^d`, we take the natural log (ln) of both:

ln(a^b) = b*ln(a) – (1)

ln(c^d) = d*ln(c) – (2)

Now, instead of comparing `a^b` and `c^d`, we can compare `b*ln(a)` and `d*ln(c)`.

Using this method, you effectively bypass the need to calculate the actual large numbers.

To achieve this in an algorithm:

1. Iterate over each line in the given text file, treating each line as a base/exponent pair.

2. For each pair (a, b), compute the value of b*ln(a).

3. Keep track of the largest value you find as you iterate through the file (initialize with the value from the first line), and when a new larger value is computed, note down the line number.

4. Return the line number that had the greatest numerical value.

Remember that Python’s (or any other language’s)log function automatically computes the natural logarithm, so you can directly use it: ln(a) = math.log(a).

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