How to prove this equation $log^2_a(x) = log_a(x^{log_ax})$?












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I depicted both these functions on the Cartesian coordinate plane and they turned out to have the same graphs. My question is: How can I get the first function from the second one doing only algebraic steps?










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    0












    $begingroup$


    I depicted both these functions on the Cartesian coordinate plane and they turned out to have the same graphs. My question is: How can I get the first function from the second one doing only algebraic steps?










    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      I depicted both these functions on the Cartesian coordinate plane and they turned out to have the same graphs. My question is: How can I get the first function from the second one doing only algebraic steps?










      share|cite|improve this question











      $endgroup$




      I depicted both these functions on the Cartesian coordinate plane and they turned out to have the same graphs. My question is: How can I get the first function from the second one doing only algebraic steps?







      functions logarithms






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      edited Dec 29 '18 at 16:21









      Bernard

      119k639112




      119k639112










      asked Dec 29 '18 at 15:58









      Vitalii PaprotskyiVitalii Paprotskyi

      6517




      6517






















          3 Answers
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          $begingroup$

          Well, by the property of logarithm, $log_a(x^{log_a x}) = log_a(x)log_a(x) = log_a^2(x)$.






          share|cite|improve this answer









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          • $begingroup$
            oh, of course, I completely forgot about it... Thanks!
            $endgroup$
            – Vitalii Paprotskyi
            Dec 29 '18 at 16:03



















          2












          $begingroup$

          Use that $$log_a{x^r}=rlog_a{x}$$ for $$x>0,a>0,aneq 1$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Recall the following property of logarithms:



            $$log_a b^c = clog_a b; quad b > 0$$



            Applying it to the right-hand side, you get



            $$log_a xcdotlog_a x = (log_a x)^2 = log^2_a x$$






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

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              3 Answers
              3






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Well, by the property of logarithm, $log_a(x^{log_a x}) = log_a(x)log_a(x) = log_a^2(x)$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                oh, of course, I completely forgot about it... Thanks!
                $endgroup$
                – Vitalii Paprotskyi
                Dec 29 '18 at 16:03
















              2












              $begingroup$

              Well, by the property of logarithm, $log_a(x^{log_a x}) = log_a(x)log_a(x) = log_a^2(x)$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                oh, of course, I completely forgot about it... Thanks!
                $endgroup$
                – Vitalii Paprotskyi
                Dec 29 '18 at 16:03














              2












              2








              2





              $begingroup$

              Well, by the property of logarithm, $log_a(x^{log_a x}) = log_a(x)log_a(x) = log_a^2(x)$.






              share|cite|improve this answer









              $endgroup$



              Well, by the property of logarithm, $log_a(x^{log_a x}) = log_a(x)log_a(x) = log_a^2(x)$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 29 '18 at 16:00









              WuestenfuxWuestenfux

              3,8311411




              3,8311411












              • $begingroup$
                oh, of course, I completely forgot about it... Thanks!
                $endgroup$
                – Vitalii Paprotskyi
                Dec 29 '18 at 16:03


















              • $begingroup$
                oh, of course, I completely forgot about it... Thanks!
                $endgroup$
                – Vitalii Paprotskyi
                Dec 29 '18 at 16:03
















              $begingroup$
              oh, of course, I completely forgot about it... Thanks!
              $endgroup$
              – Vitalii Paprotskyi
              Dec 29 '18 at 16:03




              $begingroup$
              oh, of course, I completely forgot about it... Thanks!
              $endgroup$
              – Vitalii Paprotskyi
              Dec 29 '18 at 16:03











              2












              $begingroup$

              Use that $$log_a{x^r}=rlog_a{x}$$ for $$x>0,a>0,aneq 1$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Use that $$log_a{x^r}=rlog_a{x}$$ for $$x>0,a>0,aneq 1$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Use that $$log_a{x^r}=rlog_a{x}$$ for $$x>0,a>0,aneq 1$$






                  share|cite|improve this answer









                  $endgroup$



                  Use that $$log_a{x^r}=rlog_a{x}$$ for $$x>0,a>0,aneq 1$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 29 '18 at 16:00









                  Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                  73.5k42865




                  73.5k42865























                      1












                      $begingroup$

                      Recall the following property of logarithms:



                      $$log_a b^c = clog_a b; quad b > 0$$



                      Applying it to the right-hand side, you get



                      $$log_a xcdotlog_a x = (log_a x)^2 = log^2_a x$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Recall the following property of logarithms:



                        $$log_a b^c = clog_a b; quad b > 0$$



                        Applying it to the right-hand side, you get



                        $$log_a xcdotlog_a x = (log_a x)^2 = log^2_a x$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Recall the following property of logarithms:



                          $$log_a b^c = clog_a b; quad b > 0$$



                          Applying it to the right-hand side, you get



                          $$log_a xcdotlog_a x = (log_a x)^2 = log^2_a x$$






                          share|cite|improve this answer









                          $endgroup$



                          Recall the following property of logarithms:



                          $$log_a b^c = clog_a b; quad b > 0$$



                          Applying it to the right-hand side, you get



                          $$log_a xcdotlog_a x = (log_a x)^2 = log^2_a x$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 29 '18 at 16:01









                          KM101KM101

                          5,8711423




                          5,8711423






























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