If $a^b+b^a=800$, then find $a$ and $b$.












1












$begingroup$


I've solved many of this type of equations (without steps) by noticing certain pattern in the answers.



Can someone please give the answer with proper steps.

P.S. I know the answer will be $799$ and $1$ but I'm asking for the steps.






algebra-precalculus







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I've solved many of this type of equations (without steps) by noticing certain pattern in the answers.



    Can someone please give the answer with proper steps.

    P.S. I know the answer will be $799$ and $1$ but I'm asking for the steps.






    algebra-precalculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I've solved many of this type of equations (without steps) by noticing certain pattern in the answers.



      Can someone please give the answer with proper steps.

      P.S. I know the answer will be $799$ and $1$ but I'm asking for the steps.






      algebra-precalculus







      share|cite|improve this question











      $endgroup$




      I've solved many of this type of equations (without steps) by noticing certain pattern in the answers.



      Can someone please give the answer with proper steps.

      P.S. I know the answer will be $799$ and $1$ but I'm asking for the steps.






      algebra-precalculus




      algebra-precalculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 21:44









      user376343

      3,3532825




      3,3532825










      asked Jan 8 at 17:57









      NihilusNihilus

      111




      111






















          3 Answers
          3






          active

          oldest

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          2












          $begingroup$

          If $a,bge 2$, then $a,ble 10$, because otherwise $a^b+b^a> 2^{10}>800$. So we have to check the pairs $(a,b)$ with $2le ale ble 10$, which give no solution.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            The OP did not specify integer solutions.
            $endgroup$
            – GEdgar
            Jan 8 at 18:08










          • $begingroup$
            He also did not specify anything else about $a,b$. His own answer looked so "integral".
            $endgroup$
            – Dietrich Burde
            Jan 9 at 8:58





















          2












          $begingroup$

          For any equation of the form $a^b+b^a=k$ where $k in mathbb{C}$ there is at least one solution given by: $$a=k-1$$$$b=1$$ as $(k-1)^1+1^{(k-1)}=k-1+1=k$.
          One can also swap $a,b$ to give another solution:$$a=1$$$$b=k-1$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            For any $ain (0, 799]$ there exists a real number $binmathbb{R}$ such that $a^b+b^a = 800$.



            To see why, let $f(b):=a^b+b^a$. Now $f(1) = a + 1<800$. Also, $f(b)>b^a$ and $lim_{brightarrowinfty} b^a = +infty$. These two facts plus the fact that $f$ is continuous imply that there exits a real number $b$ such that $f(b)=800$ (see IVT).



            (Actually, if $a$ is any real greater than zero and then there still exists a positive real number $b$ where a^b+b^a=800, but the proof is a little more tricky.)




            • If a=2, then b is approximately 9.47224.

            • If a=3, then b is approximately 5.82613.

            • If a=4, then b is approximately 4.37832.

            • If a=5, then b is approximately 3.49605.

            • If a=6, then b is approximately 2.91562.

            • If a=7, then b is approximately 2.52955.

            • If a=8, then b is approximately 2.2636.

            • If a=9, then b is approximately 2.07237.

            • If a=10, then b is approximately 1.92944.


            All the ordered pairs $(a,b)$ satisfying $a^b+b^a=800$ lie on the smooth curve shown below.



            enter image description here






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

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              active

              oldest

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              active

              oldest

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              2












              $begingroup$

              If $a,bge 2$, then $a,ble 10$, because otherwise $a^b+b^a> 2^{10}>800$. So we have to check the pairs $(a,b)$ with $2le ale ble 10$, which give no solution.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                The OP did not specify integer solutions.
                $endgroup$
                – GEdgar
                Jan 8 at 18:08










              • $begingroup$
                He also did not specify anything else about $a,b$. His own answer looked so "integral".
                $endgroup$
                – Dietrich Burde
                Jan 9 at 8:58


















              2












              $begingroup$

              If $a,bge 2$, then $a,ble 10$, because otherwise $a^b+b^a> 2^{10}>800$. So we have to check the pairs $(a,b)$ with $2le ale ble 10$, which give no solution.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                The OP did not specify integer solutions.
                $endgroup$
                – GEdgar
                Jan 8 at 18:08










              • $begingroup$
                He also did not specify anything else about $a,b$. His own answer looked so "integral".
                $endgroup$
                – Dietrich Burde
                Jan 9 at 8:58
















              2












              2








              2





              $begingroup$

              If $a,bge 2$, then $a,ble 10$, because otherwise $a^b+b^a> 2^{10}>800$. So we have to check the pairs $(a,b)$ with $2le ale ble 10$, which give no solution.






              share|cite|improve this answer









              $endgroup$



              If $a,bge 2$, then $a,ble 10$, because otherwise $a^b+b^a> 2^{10}>800$. So we have to check the pairs $(a,b)$ with $2le ale ble 10$, which give no solution.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 8 at 18:02









              Dietrich BurdeDietrich Burde

              78.3k64386




              78.3k64386








              • 1




                $begingroup$
                The OP did not specify integer solutions.
                $endgroup$
                – GEdgar
                Jan 8 at 18:08










              • $begingroup$
                He also did not specify anything else about $a,b$. His own answer looked so "integral".
                $endgroup$
                – Dietrich Burde
                Jan 9 at 8:58
















              • 1




                $begingroup$
                The OP did not specify integer solutions.
                $endgroup$
                – GEdgar
                Jan 8 at 18:08










              • $begingroup$
                He also did not specify anything else about $a,b$. His own answer looked so "integral".
                $endgroup$
                – Dietrich Burde
                Jan 9 at 8:58










              1




              1




              $begingroup$
              The OP did not specify integer solutions.
              $endgroup$
              – GEdgar
              Jan 8 at 18:08




              $begingroup$
              The OP did not specify integer solutions.
              $endgroup$
              – GEdgar
              Jan 8 at 18:08












              $begingroup$
              He also did not specify anything else about $a,b$. His own answer looked so "integral".
              $endgroup$
              – Dietrich Burde
              Jan 9 at 8:58






              $begingroup$
              He also did not specify anything else about $a,b$. His own answer looked so "integral".
              $endgroup$
              – Dietrich Burde
              Jan 9 at 8:58













              2












              $begingroup$

              For any equation of the form $a^b+b^a=k$ where $k in mathbb{C}$ there is at least one solution given by: $$a=k-1$$$$b=1$$ as $(k-1)^1+1^{(k-1)}=k-1+1=k$.
              One can also swap $a,b$ to give another solution:$$a=1$$$$b=k-1$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                For any equation of the form $a^b+b^a=k$ where $k in mathbb{C}$ there is at least one solution given by: $$a=k-1$$$$b=1$$ as $(k-1)^1+1^{(k-1)}=k-1+1=k$.
                One can also swap $a,b$ to give another solution:$$a=1$$$$b=k-1$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  For any equation of the form $a^b+b^a=k$ where $k in mathbb{C}$ there is at least one solution given by: $$a=k-1$$$$b=1$$ as $(k-1)^1+1^{(k-1)}=k-1+1=k$.
                  One can also swap $a,b$ to give another solution:$$a=1$$$$b=k-1$$






                  share|cite|improve this answer









                  $endgroup$



                  For any equation of the form $a^b+b^a=k$ where $k in mathbb{C}$ there is at least one solution given by: $$a=k-1$$$$b=1$$ as $(k-1)^1+1^{(k-1)}=k-1+1=k$.
                  One can also swap $a,b$ to give another solution:$$a=1$$$$b=k-1$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 18:17









                  Peter ForemanPeter Foreman

                  2616




                  2616























                      1












                      $begingroup$

                      For any $ain (0, 799]$ there exists a real number $binmathbb{R}$ such that $a^b+b^a = 800$.



                      To see why, let $f(b):=a^b+b^a$. Now $f(1) = a + 1<800$. Also, $f(b)>b^a$ and $lim_{brightarrowinfty} b^a = +infty$. These two facts plus the fact that $f$ is continuous imply that there exits a real number $b$ such that $f(b)=800$ (see IVT).



                      (Actually, if $a$ is any real greater than zero and then there still exists a positive real number $b$ where a^b+b^a=800, but the proof is a little more tricky.)




                      • If a=2, then b is approximately 9.47224.

                      • If a=3, then b is approximately 5.82613.

                      • If a=4, then b is approximately 4.37832.

                      • If a=5, then b is approximately 3.49605.

                      • If a=6, then b is approximately 2.91562.

                      • If a=7, then b is approximately 2.52955.

                      • If a=8, then b is approximately 2.2636.

                      • If a=9, then b is approximately 2.07237.

                      • If a=10, then b is approximately 1.92944.


                      All the ordered pairs $(a,b)$ satisfying $a^b+b^a=800$ lie on the smooth curve shown below.



                      enter image description here






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        For any $ain (0, 799]$ there exists a real number $binmathbb{R}$ such that $a^b+b^a = 800$.



                        To see why, let $f(b):=a^b+b^a$. Now $f(1) = a + 1<800$. Also, $f(b)>b^a$ and $lim_{brightarrowinfty} b^a = +infty$. These two facts plus the fact that $f$ is continuous imply that there exits a real number $b$ such that $f(b)=800$ (see IVT).



                        (Actually, if $a$ is any real greater than zero and then there still exists a positive real number $b$ where a^b+b^a=800, but the proof is a little more tricky.)




                        • If a=2, then b is approximately 9.47224.

                        • If a=3, then b is approximately 5.82613.

                        • If a=4, then b is approximately 4.37832.

                        • If a=5, then b is approximately 3.49605.

                        • If a=6, then b is approximately 2.91562.

                        • If a=7, then b is approximately 2.52955.

                        • If a=8, then b is approximately 2.2636.

                        • If a=9, then b is approximately 2.07237.

                        • If a=10, then b is approximately 1.92944.


                        All the ordered pairs $(a,b)$ satisfying $a^b+b^a=800$ lie on the smooth curve shown below.



                        enter image description here






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          For any $ain (0, 799]$ there exists a real number $binmathbb{R}$ such that $a^b+b^a = 800$.



                          To see why, let $f(b):=a^b+b^a$. Now $f(1) = a + 1<800$. Also, $f(b)>b^a$ and $lim_{brightarrowinfty} b^a = +infty$. These two facts plus the fact that $f$ is continuous imply that there exits a real number $b$ such that $f(b)=800$ (see IVT).



                          (Actually, if $a$ is any real greater than zero and then there still exists a positive real number $b$ where a^b+b^a=800, but the proof is a little more tricky.)




                          • If a=2, then b is approximately 9.47224.

                          • If a=3, then b is approximately 5.82613.

                          • If a=4, then b is approximately 4.37832.

                          • If a=5, then b is approximately 3.49605.

                          • If a=6, then b is approximately 2.91562.

                          • If a=7, then b is approximately 2.52955.

                          • If a=8, then b is approximately 2.2636.

                          • If a=9, then b is approximately 2.07237.

                          • If a=10, then b is approximately 1.92944.


                          All the ordered pairs $(a,b)$ satisfying $a^b+b^a=800$ lie on the smooth curve shown below.



                          enter image description here






                          share|cite|improve this answer









                          $endgroup$



                          For any $ain (0, 799]$ there exists a real number $binmathbb{R}$ such that $a^b+b^a = 800$.



                          To see why, let $f(b):=a^b+b^a$. Now $f(1) = a + 1<800$. Also, $f(b)>b^a$ and $lim_{brightarrowinfty} b^a = +infty$. These two facts plus the fact that $f$ is continuous imply that there exits a real number $b$ such that $f(b)=800$ (see IVT).



                          (Actually, if $a$ is any real greater than zero and then there still exists a positive real number $b$ where a^b+b^a=800, but the proof is a little more tricky.)




                          • If a=2, then b is approximately 9.47224.

                          • If a=3, then b is approximately 5.82613.

                          • If a=4, then b is approximately 4.37832.

                          • If a=5, then b is approximately 3.49605.

                          • If a=6, then b is approximately 2.91562.

                          • If a=7, then b is approximately 2.52955.

                          • If a=8, then b is approximately 2.2636.

                          • If a=9, then b is approximately 2.07237.

                          • If a=10, then b is approximately 1.92944.


                          All the ordered pairs $(a,b)$ satisfying $a^b+b^a=800$ lie on the smooth curve shown below.



                          enter image description here







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 8 at 18:40









                          irchansirchans

                          99239




                          99239






























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