Equation with several exponents












5












$begingroup$


Is there a way to get the result for this value of x?



x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2])


I've tried Solve, FindInstance and I did not get a solution



I tried this:



Solve[x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2]),{x}]





equation-solving







share|improve this question









$endgroup$

















    5












    $begingroup$


    Is there a way to get the result for this value of x?



    x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2])


    I've tried Solve, FindInstance and I did not get a solution



    I tried this:



    Solve[x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2]),{x}]





    equation-solving







    share|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      Is there a way to get the result for this value of x?



      x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2])


      I've tried Solve, FindInstance and I did not get a solution



      I tried this:



      Solve[x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2]),{x}]





      equation-solving







      share|improve this question









      $endgroup$




      Is there a way to get the result for this value of x?



      x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2])


      I've tried Solve, FindInstance and I did not get a solution



      I tried this:



      Solve[x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2]),{x}]





      equation-solving




      equation-solving






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 13 hours ago









      LCarvalhoLCarvalho

      5,78642986




      5,78642986






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          Try



          NSolve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]
          
(*{x->512}*)





          share|improve this answer









          $endgroup$













          • $begingroup$
            Little detail that makes all the difference. Thanks a lot!!!
            $endgroup$
            – LCarvalho
            13 hours ago








          • 4




            $begingroup$
            Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals] also works!
            $endgroup$
            – Ulrich Neumann
            13 hours ago



















          0












          $begingroup$

          This is essentially the same as Ulrich's solution but provides the motivation for restricting the domain of Solve



          eqn = x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]) // 
          
  Simplify
          

          
(* x^x^(x^(1/6)/6) == 8^(3 8^Sqrt[2]) *)


          The RHS of eqn is real



          Element[eqn[[-1]], Reals]
          

          
(* True *)


          Consequently, finding the FunctionDomain for the LHS of eqn



          const = FunctionDomain[eqn[[1]], x] // Simplify
          

          
(* x > 0 *)


          Then,



          Solve[eqn && const, x][[1]]
          

          
(* {x -> 512} *)





          share|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

            Try



            NSolve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]
            
(*{x->512}*)





            share|improve this answer









            $endgroup$













            • $begingroup$
              Little detail that makes all the difference. Thanks a lot!!!
              $endgroup$
              – LCarvalho
              13 hours ago








            • 4




              $begingroup$
              Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals] also works!
              $endgroup$
              – Ulrich Neumann
              13 hours ago
















            6












            $begingroup$

            Try



            NSolve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]
            
(*{x->512}*)





            share|improve this answer









            $endgroup$













            • $begingroup$
              Little detail that makes all the difference. Thanks a lot!!!
              $endgroup$
              – LCarvalho
              13 hours ago








            • 4




              $begingroup$
              Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals] also works!
              $endgroup$
              – Ulrich Neumann
              13 hours ago














            6












            6








            6





            $begingroup$

            Try



            NSolve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]
            
(*{x->512}*)





            share|improve this answer









            $endgroup$



            Try



            NSolve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]
            
(*{x->512}*)






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 13 hours ago









            Ulrich NeumannUlrich Neumann

            9,181516




            9,181516












            • $begingroup$
              Little detail that makes all the difference. Thanks a lot!!!
              $endgroup$
              – LCarvalho
              13 hours ago








            • 4




              $begingroup$
              Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals] also works!
              $endgroup$
              – Ulrich Neumann
              13 hours ago


















            • $begingroup$
              Little detail that makes all the difference. Thanks a lot!!!
              $endgroup$
              – LCarvalho
              13 hours ago








            • 4




              $begingroup$
              Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals] also works!
              $endgroup$
              – Ulrich Neumann
              13 hours ago
















            $begingroup$
            Little detail that makes all the difference. Thanks a lot!!!
            $endgroup$
            – LCarvalho
            13 hours ago






            $begingroup$
            Little detail that makes all the difference. Thanks a lot!!!
            $endgroup$
            – LCarvalho
            13 hours ago






            4




            4




            $begingroup$
            Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals] also works!
            $endgroup$
            – Ulrich Neumann
            13 hours ago




            $begingroup$
            Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals] also works!
            $endgroup$
            – Ulrich Neumann
            13 hours ago











            0












            $begingroup$

            This is essentially the same as Ulrich's solution but provides the motivation for restricting the domain of Solve



            eqn = x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]) // 
            
  Simplify
            

            
(* x^x^(x^(1/6)/6) == 8^(3 8^Sqrt[2]) *)


            The RHS of eqn is real



            Element[eqn[[-1]], Reals]
            

            
(* True *)


            Consequently, finding the FunctionDomain for the LHS of eqn



            const = FunctionDomain[eqn[[1]], x] // Simplify
            

            
(* x > 0 *)


            Then,



            Solve[eqn && const, x][[1]]
            

            
(* {x -> 512} *)





            share|improve this answer









            $endgroup$


















              0












              $begingroup$

              This is essentially the same as Ulrich's solution but provides the motivation for restricting the domain of Solve



              eqn = x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]) // 
              
  Simplify
              

              
(* x^x^(x^(1/6)/6) == 8^(3 8^Sqrt[2]) *)


              The RHS of eqn is real



              Element[eqn[[-1]], Reals]
              

              
(* True *)


              Consequently, finding the FunctionDomain for the LHS of eqn



              const = FunctionDomain[eqn[[1]], x] // Simplify
              

              
(* x > 0 *)


              Then,



              Solve[eqn && const, x][[1]]
              

              
(* {x -> 512} *)





              share|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                This is essentially the same as Ulrich's solution but provides the motivation for restricting the domain of Solve



                eqn = x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]) // 
                
  Simplify
                

                
(* x^x^(x^(1/6)/6) == 8^(3 8^Sqrt[2]) *)


                The RHS of eqn is real



                Element[eqn[[-1]], Reals]
                

                
(* True *)


                Consequently, finding the FunctionDomain for the LHS of eqn



                const = FunctionDomain[eqn[[1]], x] // Simplify
                

                
(* x > 0 *)


                Then,



                Solve[eqn && const, x][[1]]
                

                
(* {x -> 512} *)





                share|improve this answer









                $endgroup$



                This is essentially the same as Ulrich's solution but provides the motivation for restricting the domain of Solve



                eqn = x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]) // 
                
  Simplify
                

                
(* x^x^(x^(1/6)/6) == 8^(3 8^Sqrt[2]) *)


                The RHS of eqn is real



                Element[eqn[[-1]], Reals]
                

                
(* True *)


                Consequently, finding the FunctionDomain for the LHS of eqn



                const = FunctionDomain[eqn[[1]], x] // Simplify
                

                
(* x > 0 *)


                Then,



                Solve[eqn && const, x][[1]]
                

                
(* {x -> 512} *)






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 6 hours ago









                Bob HanlonBob Hanlon

                60.6k33597




                60.6k33597






























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