Show that $19^{31}>13^{33}$If $displaystyle x= frac{e^{3z}}{y^4}$ then $z(x,y)$?Algebra and exponentsI'm...

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Show that $19^{31}>13^{33}$


If $displaystyle x= frac{e^{3z}}{y^4}$ then $z(x,y)$?Algebra and exponentsI'm stuck with this summation problemshow a function remains positive for a given domain$b>a>0$ , how to prove that $Big(dfrac {sqrt a +sqrt b}{2}Big )^2 < dfrac 1e Big(dfrac {b^b} {a^a}Big)^{dfrac 1{b-a}}$?Simplify $(frac{x+5}{x^2-81} + frac{x+7}{x^2-18x+81}):(frac{x+3}{x-9})^2 + frac{7+x}{9+x}$Which one is bigger? Logarithms and trigonometric functionsFinding $ lim_{xrightarrow 0}frac{(1+x)^{frac{1}{x}}-e+frac{ex}{2}}{x^2}$ without using series expansion.maximum value of expression $(sqrt{-3+4x-x^2}+4)^2+(x-5)^2$Coefficient of $x^{n-3}$ in $prod^{n}_{k=1}(x-k)$













3












$begingroup$



How can i prove that $19^{31}>13^{33}$?




What I tried



$$bigg(frac{19}{13}bigg)^2=frac{361}{169}>2>1$$
then $19^{2}>13^{2}$ and $displaystyle 19^{30}>13^{30}$



How do I show it. Help me please.






algebra-precalculus exponential-function number-comparison







share|cite|improve this question











$endgroup$

















    3












    $begingroup$



    How can i prove that $19^{31}>13^{33}$?




    What I tried



    $$bigg(frac{19}{13}bigg)^2=frac{361}{169}>2>1$$
    then $19^{2}>13^{2}$ and $displaystyle 19^{30}>13^{30}$



    How do I show it. Help me please.






    algebra-precalculus exponential-function number-comparison







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$



      How can i prove that $19^{31}>13^{33}$?




      What I tried



      $$bigg(frac{19}{13}bigg)^2=frac{361}{169}>2>1$$
      then $19^{2}>13^{2}$ and $displaystyle 19^{30}>13^{30}$



      How do I show it. Help me please.






      algebra-precalculus exponential-function number-comparison







      share|cite|improve this question











      $endgroup$





      How can i prove that $19^{31}>13^{33}$?




      What I tried



      $$bigg(frac{19}{13}bigg)^2=frac{361}{169}>2>1$$
      then $19^{2}>13^{2}$ and $displaystyle 19^{30}>13^{30}$



      How do I show it. Help me please.






      algebra-precalculus exponential-function number-comparison




      algebra-precalculus exponential-function number-comparison






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 11 hours ago









      Michael Rozenberg

      107k1894198




      107k1894198










      asked 15 hours ago









      jackyjacky

      960615




      960615






















          7 Answers
          7






          active

          oldest

          votes


















          7












          $begingroup$

          Note that by the binomial theorem
          $$left(frac{19}{13}right)^{31}=left(1+frac{6}{13}right)^{31}>binom{31}{3}frac{6^3}{13^3}=frac{31cdot 30cdot 29cdot 36}{13^3}>frac{26^4}{13^3}=16cdot 13>13^2$$
          which implies that $19^{31}>13^{33}$.



          Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Solved bonus below(before seeing it posted).
            $endgroup$
            – miniparser
            14 hours ago



















          6












          $begingroup$

          $$A=19^{31}geq 19^{16}cdot18^{15} = color{red}{19^{16}cdot 9^{15}} cdot 2^{15}$$



          $$B=13^{33} = 13cdot 169^{16}leq 13cdot 171^{16} = 13 cdot color{red}{19^{16}cdot 9^{15}}cdot 9$$



          Clearly $$2^{15}geq 13cdot 9implies A>B$$






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
            $x ln(19) > (x+2) ln(13)$
            and finally
            $ x > 2 frac{ln(13)}{ln(frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.






            share|cite|improve this answer









            $endgroup$





















              2












              $begingroup$

              By your work
              $$frac{19^2}{13^2}>2,$$ which says
              $$left(frac{19}{13}right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!



              By this way we can prove that
              $$19^{31}>13^{35}.$$






              share|cite|improve this answer











              $endgroup$





















                1












                $begingroup$

                $19^{30}=(19^3)^{10}=6859^{10}gt 13^{30}=(13^3)^{10}=2197^{10}$



                $6859gt 3times 2197$



                $6859^{10}gt (3times 2197)^{10}=3^{10}times 2197^{10}=3^3times 3^3times 3^3times 3times 2197^{10}=27^3times 3times 2197^{10}$



                $27^3times 3times 2197^{10}gt 13^3times 2197^{10}$ already without even multiplying by $19.$






                share|cite|improve this answer









                $endgroup$





















                  1












                  $begingroup$

                  Well, using what you've got so far:



                  $(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{30} > 2^{15}$ so $19^{30} > 2^{15}*13^{30}$.



                  $19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? over >} 13^{30}*13^3 = 13^{33}$. If we can show $19*(2^{15}) > 13^3$ we will be done.



                  =====



                  For what it's worth one way I might do it if it were up to me is to note.



                  $19 > 13$ so $19^k > 13^k$ For $19^k > 13^{k+2} = 13^k *13^2$ we need $(frac {19}{13})^k > 169$.



                  $(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{16} > 2^8 = 256> 169$.



                  So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.






                  share|cite|improve this answer











                  $endgroup$





















                    1












                    $begingroup$

                    $left[dfrac{19}{13}right]^{large 4}!!!=!left[1!+!dfrac{6}{13}right]^{large 4}!!>1!+!4left[dfrac{6}{13}right]!>2,overset{( ,)^{LARGE 8}}Longrightarrow, left[dfrac{19}{13}right]^{large 32}!!!>2^{large 8} > 19cdot 13$






                    share|cite|improve this answer









                    $endgroup$













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                      7 Answers
                      7






                      active

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                      7 Answers
                      7






                      active

                      oldest

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                      active

                      oldest

                      votes






                      active

                      oldest

                      votes









                      7












                      $begingroup$

                      Note that by the binomial theorem
                      $$left(frac{19}{13}right)^{31}=left(1+frac{6}{13}right)^{31}>binom{31}{3}frac{6^3}{13^3}=frac{31cdot 30cdot 29cdot 36}{13^3}>frac{26^4}{13^3}=16cdot 13>13^2$$
                      which implies that $19^{31}>13^{33}$.



                      Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Solved bonus below(before seeing it posted).
                        $endgroup$
                        – miniparser
                        14 hours ago
















                      7












                      $begingroup$

                      Note that by the binomial theorem
                      $$left(frac{19}{13}right)^{31}=left(1+frac{6}{13}right)^{31}>binom{31}{3}frac{6^3}{13^3}=frac{31cdot 30cdot 29cdot 36}{13^3}>frac{26^4}{13^3}=16cdot 13>13^2$$
                      which implies that $19^{31}>13^{33}$.



                      Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Solved bonus below(before seeing it posted).
                        $endgroup$
                        – miniparser
                        14 hours ago














                      7












                      7








                      7





                      $begingroup$

                      Note that by the binomial theorem
                      $$left(frac{19}{13}right)^{31}=left(1+frac{6}{13}right)^{31}>binom{31}{3}frac{6^3}{13^3}=frac{31cdot 30cdot 29cdot 36}{13^3}>frac{26^4}{13^3}=16cdot 13>13^2$$
                      which implies that $19^{31}>13^{33}$.



                      Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?






                      share|cite|improve this answer











                      $endgroup$



                      Note that by the binomial theorem
                      $$left(frac{19}{13}right)^{31}=left(1+frac{6}{13}right)^{31}>binom{31}{3}frac{6^3}{13^3}=frac{31cdot 30cdot 29cdot 36}{13^3}>frac{26^4}{13^3}=16cdot 13>13^2$$
                      which implies that $19^{31}>13^{33}$.



                      Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 14 hours ago

























                      answered 15 hours ago









                      Robert ZRobert Z

                      99.8k1068140




                      99.8k1068140












                      • $begingroup$
                        Solved bonus below(before seeing it posted).
                        $endgroup$
                        – miniparser
                        14 hours ago


















                      • $begingroup$
                        Solved bonus below(before seeing it posted).
                        $endgroup$
                        – miniparser
                        14 hours ago
















                      $begingroup$
                      Solved bonus below(before seeing it posted).
                      $endgroup$
                      – miniparser
                      14 hours ago




                      $begingroup$
                      Solved bonus below(before seeing it posted).
                      $endgroup$
                      – miniparser
                      14 hours ago











                      6












                      $begingroup$

                      $$A=19^{31}geq 19^{16}cdot18^{15} = color{red}{19^{16}cdot 9^{15}} cdot 2^{15}$$



                      $$B=13^{33} = 13cdot 169^{16}leq 13cdot 171^{16} = 13 cdot color{red}{19^{16}cdot 9^{15}}cdot 9$$



                      Clearly $$2^{15}geq 13cdot 9implies A>B$$






                      share|cite|improve this answer









                      $endgroup$


















                        6












                        $begingroup$

                        $$A=19^{31}geq 19^{16}cdot18^{15} = color{red}{19^{16}cdot 9^{15}} cdot 2^{15}$$



                        $$B=13^{33} = 13cdot 169^{16}leq 13cdot 171^{16} = 13 cdot color{red}{19^{16}cdot 9^{15}}cdot 9$$



                        Clearly $$2^{15}geq 13cdot 9implies A>B$$






                        share|cite|improve this answer









                        $endgroup$
















                          6












                          6








                          6





                          $begingroup$

                          $$A=19^{31}geq 19^{16}cdot18^{15} = color{red}{19^{16}cdot 9^{15}} cdot 2^{15}$$



                          $$B=13^{33} = 13cdot 169^{16}leq 13cdot 171^{16} = 13 cdot color{red}{19^{16}cdot 9^{15}}cdot 9$$



                          Clearly $$2^{15}geq 13cdot 9implies A>B$$






                          share|cite|improve this answer









                          $endgroup$



                          $$A=19^{31}geq 19^{16}cdot18^{15} = color{red}{19^{16}cdot 9^{15}} cdot 2^{15}$$



                          $$B=13^{33} = 13cdot 169^{16}leq 13cdot 171^{16} = 13 cdot color{red}{19^{16}cdot 9^{15}}cdot 9$$



                          Clearly $$2^{15}geq 13cdot 9implies A>B$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 15 hours ago









                          greedoidgreedoid

                          46.1k1160117




                          46.1k1160117























                              4












                              $begingroup$

                              Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
                              $x ln(19) > (x+2) ln(13)$
                              and finally
                              $ x > 2 frac{ln(13)}{ln(frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.






                              share|cite|improve this answer









                              $endgroup$


















                                4












                                $begingroup$

                                Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
                                $x ln(19) > (x+2) ln(13)$
                                and finally
                                $ x > 2 frac{ln(13)}{ln(frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.






                                share|cite|improve this answer









                                $endgroup$
















                                  4












                                  4








                                  4





                                  $begingroup$

                                  Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
                                  $x ln(19) > (x+2) ln(13)$
                                  and finally
                                  $ x > 2 frac{ln(13)}{ln(frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.






                                  share|cite|improve this answer









                                  $endgroup$



                                  Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
                                  $x ln(19) > (x+2) ln(13)$
                                  and finally
                                  $ x > 2 frac{ln(13)}{ln(frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered 15 hours ago









                                  Gâteau-GalloisGâteau-Gallois

                                  410213




                                  410213























                                      2












                                      $begingroup$

                                      By your work
                                      $$frac{19^2}{13^2}>2,$$ which says
                                      $$left(frac{19}{13}right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!



                                      By this way we can prove that
                                      $$19^{31}>13^{35}.$$






                                      share|cite|improve this answer











                                      $endgroup$


















                                        2












                                        $begingroup$

                                        By your work
                                        $$frac{19^2}{13^2}>2,$$ which says
                                        $$left(frac{19}{13}right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!



                                        By this way we can prove that
                                        $$19^{31}>13^{35}.$$






                                        share|cite|improve this answer











                                        $endgroup$
















                                          2












                                          2








                                          2





                                          $begingroup$

                                          By your work
                                          $$frac{19^2}{13^2}>2,$$ which says
                                          $$left(frac{19}{13}right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!



                                          By this way we can prove that
                                          $$19^{31}>13^{35}.$$






                                          share|cite|improve this answer











                                          $endgroup$



                                          By your work
                                          $$frac{19^2}{13^2}>2,$$ which says
                                          $$left(frac{19}{13}right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!



                                          By this way we can prove that
                                          $$19^{31}>13^{35}.$$







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          edited 11 hours ago

























                                          answered 11 hours ago









                                          Michael RozenbergMichael Rozenberg

                                          107k1894198




                                          107k1894198























                                              1












                                              $begingroup$

                                              $19^{30}=(19^3)^{10}=6859^{10}gt 13^{30}=(13^3)^{10}=2197^{10}$



                                              $6859gt 3times 2197$



                                              $6859^{10}gt (3times 2197)^{10}=3^{10}times 2197^{10}=3^3times 3^3times 3^3times 3times 2197^{10}=27^3times 3times 2197^{10}$



                                              $27^3times 3times 2197^{10}gt 13^3times 2197^{10}$ already without even multiplying by $19.$






                                              share|cite|improve this answer









                                              $endgroup$


















                                                1












                                                $begingroup$

                                                $19^{30}=(19^3)^{10}=6859^{10}gt 13^{30}=(13^3)^{10}=2197^{10}$



                                                $6859gt 3times 2197$



                                                $6859^{10}gt (3times 2197)^{10}=3^{10}times 2197^{10}=3^3times 3^3times 3^3times 3times 2197^{10}=27^3times 3times 2197^{10}$



                                                $27^3times 3times 2197^{10}gt 13^3times 2197^{10}$ already without even multiplying by $19.$






                                                share|cite|improve this answer









                                                $endgroup$
















                                                  1












                                                  1








                                                  1





                                                  $begingroup$

                                                  $19^{30}=(19^3)^{10}=6859^{10}gt 13^{30}=(13^3)^{10}=2197^{10}$



                                                  $6859gt 3times 2197$



                                                  $6859^{10}gt (3times 2197)^{10}=3^{10}times 2197^{10}=3^3times 3^3times 3^3times 3times 2197^{10}=27^3times 3times 2197^{10}$



                                                  $27^3times 3times 2197^{10}gt 13^3times 2197^{10}$ already without even multiplying by $19.$






                                                  share|cite|improve this answer









                                                  $endgroup$



                                                  $19^{30}=(19^3)^{10}=6859^{10}gt 13^{30}=(13^3)^{10}=2197^{10}$



                                                  $6859gt 3times 2197$



                                                  $6859^{10}gt (3times 2197)^{10}=3^{10}times 2197^{10}=3^3times 3^3times 3^3times 3times 2197^{10}=27^3times 3times 2197^{10}$



                                                  $27^3times 3times 2197^{10}gt 13^3times 2197^{10}$ already without even multiplying by $19.$







                                                  share|cite|improve this answer












                                                  share|cite|improve this answer



                                                  share|cite|improve this answer










                                                  answered 14 hours ago









                                                  miniparserminiparser

                                                  7871616




                                                  7871616























                                                      1












                                                      $begingroup$

                                                      Well, using what you've got so far:



                                                      $(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{30} > 2^{15}$ so $19^{30} > 2^{15}*13^{30}$.



                                                      $19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? over >} 13^{30}*13^3 = 13^{33}$. If we can show $19*(2^{15}) > 13^3$ we will be done.



                                                      =====



                                                      For what it's worth one way I might do it if it were up to me is to note.



                                                      $19 > 13$ so $19^k > 13^k$ For $19^k > 13^{k+2} = 13^k *13^2$ we need $(frac {19}{13})^k > 169$.



                                                      $(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{16} > 2^8 = 256> 169$.



                                                      So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.






                                                      share|cite|improve this answer











                                                      $endgroup$


















                                                        1












                                                        $begingroup$

                                                        Well, using what you've got so far:



                                                        $(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{30} > 2^{15}$ so $19^{30} > 2^{15}*13^{30}$.



                                                        $19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? over >} 13^{30}*13^3 = 13^{33}$. If we can show $19*(2^{15}) > 13^3$ we will be done.



                                                        =====



                                                        For what it's worth one way I might do it if it were up to me is to note.



                                                        $19 > 13$ so $19^k > 13^k$ For $19^k > 13^{k+2} = 13^k *13^2$ we need $(frac {19}{13})^k > 169$.



                                                        $(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{16} > 2^8 = 256> 169$.



                                                        So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.






                                                        share|cite|improve this answer











                                                        $endgroup$
















                                                          1












                                                          1








                                                          1





                                                          $begingroup$

                                                          Well, using what you've got so far:



                                                          $(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{30} > 2^{15}$ so $19^{30} > 2^{15}*13^{30}$.



                                                          $19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? over >} 13^{30}*13^3 = 13^{33}$. If we can show $19*(2^{15}) > 13^3$ we will be done.



                                                          =====



                                                          For what it's worth one way I might do it if it were up to me is to note.



                                                          $19 > 13$ so $19^k > 13^k$ For $19^k > 13^{k+2} = 13^k *13^2$ we need $(frac {19}{13})^k > 169$.



                                                          $(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{16} > 2^8 = 256> 169$.



                                                          So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.






                                                          share|cite|improve this answer











                                                          $endgroup$



                                                          Well, using what you've got so far:



                                                          $(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{30} > 2^{15}$ so $19^{30} > 2^{15}*13^{30}$.



                                                          $19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? over >} 13^{30}*13^3 = 13^{33}$. If we can show $19*(2^{15}) > 13^3$ we will be done.



                                                          =====



                                                          For what it's worth one way I might do it if it were up to me is to note.



                                                          $19 > 13$ so $19^k > 13^k$ For $19^k > 13^{k+2} = 13^k *13^2$ we need $(frac {19}{13})^k > 169$.



                                                          $(frac {19}{13})^2 > 2$ so $(frac {19}{13})^{16} > 2^8 = 256> 169$.



                                                          So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.







                                                          share|cite|improve this answer














                                                          share|cite|improve this answer



                                                          share|cite|improve this answer








                                                          edited 10 hours ago

























                                                          answered 10 hours ago









                                                          fleabloodfleablood

                                                          71.9k22687




                                                          71.9k22687























                                                              1












                                                              $begingroup$

                                                              $left[dfrac{19}{13}right]^{large 4}!!!=!left[1!+!dfrac{6}{13}right]^{large 4}!!>1!+!4left[dfrac{6}{13}right]!>2,overset{( ,)^{LARGE 8}}Longrightarrow, left[dfrac{19}{13}right]^{large 32}!!!>2^{large 8} > 19cdot 13$






                                                              share|cite|improve this answer









                                                              $endgroup$


















                                                                1












                                                                $begingroup$

                                                                $left[dfrac{19}{13}right]^{large 4}!!!=!left[1!+!dfrac{6}{13}right]^{large 4}!!>1!+!4left[dfrac{6}{13}right]!>2,overset{( ,)^{LARGE 8}}Longrightarrow, left[dfrac{19}{13}right]^{large 32}!!!>2^{large 8} > 19cdot 13$






                                                                share|cite|improve this answer









                                                                $endgroup$
















                                                                  1












                                                                  1








                                                                  1





                                                                  $begingroup$

                                                                  $left[dfrac{19}{13}right]^{large 4}!!!=!left[1!+!dfrac{6}{13}right]^{large 4}!!>1!+!4left[dfrac{6}{13}right]!>2,overset{( ,)^{LARGE 8}}Longrightarrow, left[dfrac{19}{13}right]^{large 32}!!!>2^{large 8} > 19cdot 13$






                                                                  share|cite|improve this answer









                                                                  $endgroup$



                                                                  $left[dfrac{19}{13}right]^{large 4}!!!=!left[1!+!dfrac{6}{13}right]^{large 4}!!>1!+!4left[dfrac{6}{13}right]!>2,overset{( ,)^{LARGE 8}}Longrightarrow, left[dfrac{19}{13}right]^{large 32}!!!>2^{large 8} > 19cdot 13$







                                                                  share|cite|improve this answer












                                                                  share|cite|improve this answer



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                                                                  answered 7 hours ago









                                                                  Bill DubuqueBill Dubuque

                                                                  212k29195650




                                                                  212k29195650






























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