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First, let us look for some similarities between 17 and 31 which we may be able to use to our advantage. Both numbers are just 1 away from a power of 2:

$17=16+1$ and $31=32-1$ .

Using this similarity...

${ 16 }^{ 31 }={ ({ 2 }^{ 4 }) }^{ 31 }={ 2 }^{ 124 }$ . Clearly, ${ 17 }^{ 31 }{ >16 }^{ 31 }$ so ${ 17 }^{ 31 }>{ 2 }^{ 124 }$ .

${ 32 }^{ 17 }={ ({ 2 }^{ 5 }) }^{ 17 }={ 2 }^{ 85 }$ . Clearly, ${ 31 }^{ 17 }{ <32 }^{ 17 }$ so ${ 31 }^{ 17 }<{ 2 }^{ 85 }$ .

Since ${ 2 }^{ 124 }>{ 2 }^{ 85 }$ , we can definitely say that ${ 17 }^{ 31 }>{ 31 }^{ 17 }$ .

For the number closer to $e$ , $x^{\frac{1}{x}}$ is greater. For proof, check the function $f(x) = \displaystyle x^{\frac{1}{x}}$ . It achieves it's maximum at $x=e$ . This gives $\sqrt[17]{17} > \sqrt[31]{31}$ . Result follows.

17>16 which is 2^4 therefore 17^31>2^(431)>2^124 31<32 which is 2^5 therefore 31^17<2^(517)<2^85 Hence 31^17 is the answer

i just picked smaller numbers like 2^5 and 5^2 ,reasoned out the circumstances which makes the first instance(2^5) greater,and applied it to the question. it worked!!

Take log values.

Clearly 31log17>17log31

We know that the 17^31 is of the order 10^31. 3^17is of the order 10^17 and we have almost a factor of 2 (31/17) to worry about. 2^31 is of the order 10^9.

Usually the smaller number to the power of a bigger number is greater than the bigger number to the power of a smaller number. ex: 3 to the power of 2 and 2 to the power of 3 ( 2 to the power of 3 is greater)

Using small number for instace take:

3^2 and 2^3, we realize that 3^2 = 9 and 2^3 = 8

therefore this implies that 31^17 > 17^31

this is the true power of exponential functions, small no. with a higher power grows faster than a large number with a smaller power. I know this logic and got the answer without even thinking, as difference between 31 and 17 is large.....

17^31= 17^(17+14) = (17^17).(17^14) = (17^17).(14+3)^14 = (17^17).(14^14).[1+(3/14)]^14 = (17^17).(14^14).(approx 5/4)^14 and 31^17 = (17+14)^17 = (17^17).[1+(14/17)]^17 = (17^17). (approx 2)^17

Clearly the term 14^14 will take a major lead over 2^17 and hence 17^31 will be greater.

Let take y=17^31 and y'=31^17 Now just take log both side to get logy=31log17 and logy'=17log31 Now you can just use the log properties and approximation to know that whose bigger? logy or logy' ? you will just get an ans.. QED

it's simple calculate its ten power using logarthims i.e17^31 has a ten power 38 whereas 31^17 has ten power 25

we know 17^31 > 16^31=2^124 and 31^17< 32^17= 2^85 seeing above we got 17^31>31^17

Take the natural logarithm of both sides, to get 17 log(31) & 31 log(17). Since logarithm is a monotonically increasing function, the relationship (greater, lesser or equal) between the two sides is preserved in the transformation. Now, cross-multiply to get $\frac{log17}{17}$ & $\frac{log31}{31}$ .

Consider the function $\frac{log x}{x}$ . Differentiating this gives $\frac{1-log(x)}{x^2}$ , which is strictly negative for all x>e (Denominator is a square and always positive, numerator is negative since log(x)>1 for all x>e). This means that the function is monotonically decreasing, i.e., $\frac{log 17}{17} > \frac{log 31}{31}$ . Thus, 17 log(31)<31 log(17) and so, $17^{31} > 31^{17}$ .

assume function x^1/x and differentiate it to get where the function is increasing or decreasing. Now you can know the relation between 17^1/17 and 31^1/31 which is 17^1/17> 31^1/31. now raise both sides to 17*31. Since the numbers are greater than 1, inequality holds :)

consider 31^17. taking log will give 17log31 now, 17log31< 17log34 17log34= 34log17 34log17 < 31log17..... hence 17log31 < 31log17 therefore 31^17 < 17^31

17^31 obviously17^31 obviously as it can be written as (1.7 ×10)^31 so it means 1.7^31× 10^31..... nd 31^17 can be written as (3.1×10)^17 therefore = 3.1^17×10^17 now as 10^31>10^17 therefore mutually it is directly stated 17^31> 31^17 ....

As 3.1>1.7 bt nos of zeroes on side 17^31 will always greater than 31^17

Dats all ....

as we know that 17=16+1 and 31=32-1 solving it correctly we will get
31^17<2^85 and 17^31>2^124 so 17^ 31 is greater