Let's simplify!
We can't uniform the bases, but it's obvious that exponents can be simplified (they are all multiples of 6).

First,
\(2^{120} = (2^{20})^6\)
\(3^{72}=(3^{12})^6\)
\(17^{30}=(17^5)^6\)

ok, this didn't help much, but let's try to compare only 2 at a time:

\(2^{120} = (2^{20})^6 = (2^{5})^{24} = 32^{24}\)
\(3^{72}=(3^{12})^6 = (3^{3})^{24} = 27^{24}\)

Clearly, I > II

Great! Now, lets compare first and third (120 and 30 are easier).

\(2^{120} = (2^{4})^{30} = 16^{30}\), clearly < \(17^{30}\)

Hence, III > I > II. Answer:

Show Spoiler

C

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We need to compare \(2^{12}\), \(3^{72}\) and \(17^{30}\)

The exponents 120, 72 and 30 have 6 as common factor.

rewriting, \((2^{20}){6}\), \((2^{12}){6}\), \((17^{5}){6}\)

the last term: \((2^4 +1 )^{5}){6}\)
\((2^{20} + 5 *2^{14} +......){6}\)
This is definitely greater than the first term..Second term is the smallest.

Hence, III > I > II

in such cases we should try to equate similar powers

2^120 = (2^8)^15
17^30 = (17^2)^15

2^8<17^2, therefore, 2^120<17^30

now compare 2^120 and 3^72
2^120= (2^10)^12
3^72= (3^6)^12

2^10>3^6 therefore 3^72<2^120

Finally, 3^72<2^120<17^30

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