MHB Comparison between two numbers

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To compare the two expressions, 1016^{11}·3016^{31} and 2016^{42}, one approach is to use logarithms to simplify the calculations. By taking the logarithm of both sides, the problem can be reduced to comparing the sums of the logarithms of the bases multiplied by their respective exponents. The first expression can be expressed as log(1016^{11}) + log(3016^{31}), while the second becomes log(2016^{42}). After calculating these logarithmic values, it becomes clear which expression is larger. The comparison ultimately reveals that 2016^{42} is the larger number.
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Determine, without the help of calculator, which of the following is bigger:

$$1016^{11}\cdot 3016^{31}$$ versus $$2016^{42}$$
 
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We know $3^6(= 729) > 2^9(=512)$ hence $3^{30} > 2^{45}$
Or $3^{31} > 3 * 2^{45} > 24 * 2^{42}$
so $(1000)^{11} * {3000}^{31} > 24 * {2000}^{42}$
Hence $(1016)^{11} * {3016}^{31} > 24 * {2000}^{42}\cdots(1)$
Now $2016^{42} = 2000^{42}( 1+ \frac{1}{125})^{42} < 2000^{42}( 1+ \frac{1}{125})^{125} < 2000^{42} * e < 2000^{42} *3$
from above and (1) we have $(1016)^{11} * {3016}^{31} > {2016}^{42}\cdots(3)$
Hence $(1016)^{11} * {3016}^{31}$ is larger
 
Awesome! Thanks kaliprasad for participating!(Cool)

My solution:

I first assume $$1016^{11}\cdot 3016^{31}\gt 2016^{42}$$.

From the prime factorization of $$1016=2^3\cdot 127,\,2016=2^5\cdot 3^2\cdot 7,\,3016=2^3\cdot 13\cdot 29$$

I need to prove

$$(2^3\cdot 127)^{11}\cdot (2^3\cdot 13\cdot 29)^{31}\gt (2^5\cdot 3^2\cdot 7)^{42}$$

Simplify the above we get:

$$2^{33}\cdot 127^{11}\cdot 2^{93}\cdot 13^{31}\cdot 29^{31}\gt 2^{210}\cdot 3^{84}\cdot 7^{42}$$

$$127^{11}\cdot 13^{31}\cdot 29^{31}\gt 2^{210-33-93}\cdot 3^{84}\cdot 7^{42}$$

$$127^{11}\cdot 13^{31}\cdot 29^{31}\gt 2^{84}\cdot 3^{84}\cdot 7^{42}$$

$$127^{11}\cdot 377^{31}\gt 6^{84}\cdot 7^{42}$$*

Observe that:

$7^2=49,\,7^3=343\implies 373\gt 7^3$

Therefore $373^{14}\gt (7^3)^{14}$ which is $373^{14}\gt 7^{42}$

At this point, if we can prove

$$127^{11}\cdot 377^{17}\gt 6^{84}$$

Then we can conclude $$1016^{11}\cdot 3016^{31}\gt 2016^{42}$$ is correct.

Observe again that

$127\cdot 377=47879\gt 6^6=46656$

Taking 14th power on both sides of the inequality we get:

$(127\cdot 377)^{14}\gt (6^6)^{14}$

$$127^{14}\cdot 377^{14}\gt 6^{84}$$

It's obvious that

$$127^{11}\cdot 377^{17}\gt 127^{14}\cdot 377^{14}$$ is true, therefore, our assumption has been justified, therefore $$1016^{11}\cdot 3016^{31}$$ is bigger than $$2016^{42}$$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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