Question 1045312
.
Without using a calculator or technology, determine which is bigger, 

{{{sqrt(2017)^(sqrt(2016))}}} or {{{sqrt(2016)^(sqrt(2017))}}}.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
Let x = {{{sqrt(2017)^(sqrt(2016))}}}  and  y = {{{sqrt(2016)^(sqrt(2017))}}}.


Then  

ln(x) = {{{sqrt(2016)*ln(sqrt(2017))}}} = {{{(1/2)*sqrt(2016)*ln(2017)}}}    and  

ln(y) = {{{sqrt(2017)*ln(sqrt(2016))}}} = {{{(1/2)*sqrt(2017)*ln(2016)}}}.


Then

{{{ln(x)/ln(y)}}} = {{{(sqrt(2016)*ln(2017))/(sqrt(2017)*ln(2016))}}} = {{{((sqrt(2016)/ln(2016)))/((sqrt(2017)/ln(2017)))}}}.

But  {{{sqrt(x)/ln(x)}}} is monotonically increasing function, as everybody knows (or everybody should know who studies/studied Calculus).

Therefore,  {{{ln(x)/ln(y)}}} = {{{((sqrt(2016)/ln(2016)))/((sqrt(2017)/ln(2017)))}}} < 1.


It implies that  ln(x) < ln(y).


Hence,  x = {{{sqrt(2017)^(sqrt(2016))}}}  <  y = {{{sqrt(2016)^(sqrt(2017))}}}.


<U>Answer</U>.  {{{sqrt(2017)^(sqrt(2016))}}}  <  {{{sqrt(2016)^(sqrt(2017))}}}.
</pre>

By the way, there is a classic question: which number is bigger (or larger, or greater):


{{{e^pi}}}  or  {{{pi^e}}}.


The way to solve it is the same.



For more elementary inequalities see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Inequalities/Which-number-is-greater-Comparing-magnitude-of-irrational-numbers.lesson>What number is greater? Comparing magnitude of irrational numbers</A> 

in this site.



(one more time) By the way, you have free of charge online textbook in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.