Find range of $x$ if $log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)leq1$












2












$begingroup$



If $log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)leq1$, then $x$ lies in _______




My Attempt



$$
log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)=log_5bigg(6+dfrac{2}{x}bigg)-log_{5}bigg(1+dfrac{x}{10}bigg)leq1\
log_5frac{(6x+2)10}{x(10+x)}leq1impliesfrac{(6x+2)10}{x(10+x)}leq5\
frac{4(3x+1)}{x^2+10x}leq1\
implies 12x+4leq x^2+10xquadtext{or}quad12x+4>x^2+10x\
x^2-2x-4geq0quadtext{or}quad x^2-2x-4<0implies xinmathcal{R}
$$



My reference gives the solution $(-infty,1-sqrt{5})cup(1+sqrt{5},infty)$, what is going wrong here ?






inequality logarithms







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    2












    $begingroup$



    If $log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)leq1$, then $x$ lies in _______




    My Attempt



    $$
    log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)=log_5bigg(6+dfrac{2}{x}bigg)-log_{5}bigg(1+dfrac{x}{10}bigg)leq1\
    log_5frac{(6x+2)10}{x(10+x)}leq1impliesfrac{(6x+2)10}{x(10+x)}leq5\
    frac{4(3x+1)}{x^2+10x}leq1\
    implies 12x+4leq x^2+10xquadtext{or}quad12x+4>x^2+10x\
    x^2-2x-4geq0quadtext{or}quad x^2-2x-4<0implies xinmathcal{R}
    $$



    My reference gives the solution $(-infty,1-sqrt{5})cup(1+sqrt{5},infty)$, what is going wrong here ?






    inequality logarithms







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      If $log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)leq1$, then $x$ lies in _______




      My Attempt



      $$
      log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)=log_5bigg(6+dfrac{2}{x}bigg)-log_{5}bigg(1+dfrac{x}{10}bigg)leq1\
      log_5frac{(6x+2)10}{x(10+x)}leq1impliesfrac{(6x+2)10}{x(10+x)}leq5\
      frac{4(3x+1)}{x^2+10x}leq1\
      implies 12x+4leq x^2+10xquadtext{or}quad12x+4>x^2+10x\
      x^2-2x-4geq0quadtext{or}quad x^2-2x-4<0implies xinmathcal{R}
      $$



      My reference gives the solution $(-infty,1-sqrt{5})cup(1+sqrt{5},infty)$, what is going wrong here ?






      inequality logarithms







      share|cite|improve this question











      $endgroup$





      If $log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)leq1$, then $x$ lies in _______




      My Attempt



      $$
      log_5bigg(6+dfrac{2}{x}bigg)+log_{1/5}bigg(1+dfrac{x}{10}bigg)=log_5bigg(6+dfrac{2}{x}bigg)-log_{5}bigg(1+dfrac{x}{10}bigg)leq1\
      log_5frac{(6x+2)10}{x(10+x)}leq1impliesfrac{(6x+2)10}{x(10+x)}leq5\
      frac{4(3x+1)}{x^2+10x}leq1\
      implies 12x+4leq x^2+10xquadtext{or}quad12x+4>x^2+10x\
      x^2-2x-4geq0quadtext{or}quad x^2-2x-4<0implies xinmathcal{R}
      $$



      My reference gives the solution $(-infty,1-sqrt{5})cup(1+sqrt{5},infty)$, what is going wrong here ?






      inequality logarithms




      inequality logarithms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 23 at 19:14







      ss1729

















      asked Jan 23 at 14:21









      ss1729ss1729

      2,00311024




      2,00311024






















          3 Answers
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          2












          $begingroup$

          You forgot to check when $$6+dfrac{2}{x}>0$$ and $$1+dfrac{x}{10}>0$$



          is true!






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            ohh i can't believe i did not see the domain of log got to be greater than zero.
            $endgroup$
            – ss1729
            Jan 23 at 19:18



















          1












          $begingroup$

          By your work we need to solve
          $$frac{x^2-2x-4}{x(x+10)}geq0$$ and the domain gives $x>0$ or $-10<x<-frac{1}{3}.$



          The first by the interval's method gives
          $$1-sqrt5leq x<0$$ or $$xgeq1+sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
          $$left[1-sqrt5,-frac{1}{3}right)cup[1+sqrt5,+infty).$$






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Writing your inequality in the form $$frac{lnleft(6+frac{2}{x}right)}{ln(5)}+frac{lnleft(1+frac{x}{10}right)}{-ln(5)}le 1$$ we get the inequalities
            $$6+frac{2}{x}>0$$ and $$1+frac{x}{10}>0$$ and
            $$frac{6+frac{2}{x}}{1+frac{x}{10}}le 5$$ we get
            $$1-sqrt{5}le x<-frac{1}{3}$$ or $$xgeq 1+sqrt{5}$$






            share|cite|improve this answer











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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              You forgot to check when $$6+dfrac{2}{x}>0$$ and $$1+dfrac{x}{10}>0$$



              is true!






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                ohh i can't believe i did not see the domain of log got to be greater than zero.
                $endgroup$
                – ss1729
                Jan 23 at 19:18
















              2












              $begingroup$

              You forgot to check when $$6+dfrac{2}{x}>0$$ and $$1+dfrac{x}{10}>0$$



              is true!






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                ohh i can't believe i did not see the domain of log got to be greater than zero.
                $endgroup$
                – ss1729
                Jan 23 at 19:18














              2












              2








              2





              $begingroup$

              You forgot to check when $$6+dfrac{2}{x}>0$$ and $$1+dfrac{x}{10}>0$$



              is true!






              share|cite|improve this answer









              $endgroup$



              You forgot to check when $$6+dfrac{2}{x}>0$$ and $$1+dfrac{x}{10}>0$$



              is true!







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 23 at 14:27









              Maria MazurMaria Mazur

              46.7k1260120




              46.7k1260120












              • $begingroup$
                ohh i can't believe i did not see the domain of log got to be greater than zero.
                $endgroup$
                – ss1729
                Jan 23 at 19:18


















              • $begingroup$
                ohh i can't believe i did not see the domain of log got to be greater than zero.
                $endgroup$
                – ss1729
                Jan 23 at 19:18
















              $begingroup$
              ohh i can't believe i did not see the domain of log got to be greater than zero.
              $endgroup$
              – ss1729
              Jan 23 at 19:18




              $begingroup$
              ohh i can't believe i did not see the domain of log got to be greater than zero.
              $endgroup$
              – ss1729
              Jan 23 at 19:18











              1












              $begingroup$

              By your work we need to solve
              $$frac{x^2-2x-4}{x(x+10)}geq0$$ and the domain gives $x>0$ or $-10<x<-frac{1}{3}.$



              The first by the interval's method gives
              $$1-sqrt5leq x<0$$ or $$xgeq1+sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
              $$left[1-sqrt5,-frac{1}{3}right)cup[1+sqrt5,+infty).$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                By your work we need to solve
                $$frac{x^2-2x-4}{x(x+10)}geq0$$ and the domain gives $x>0$ or $-10<x<-frac{1}{3}.$



                The first by the interval's method gives
                $$1-sqrt5leq x<0$$ or $$xgeq1+sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
                $$left[1-sqrt5,-frac{1}{3}right)cup[1+sqrt5,+infty).$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  By your work we need to solve
                  $$frac{x^2-2x-4}{x(x+10)}geq0$$ and the domain gives $x>0$ or $-10<x<-frac{1}{3}.$



                  The first by the interval's method gives
                  $$1-sqrt5leq x<0$$ or $$xgeq1+sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
                  $$left[1-sqrt5,-frac{1}{3}right)cup[1+sqrt5,+infty).$$






                  share|cite|improve this answer











                  $endgroup$



                  By your work we need to solve
                  $$frac{x^2-2x-4}{x(x+10)}geq0$$ and the domain gives $x>0$ or $-10<x<-frac{1}{3}.$



                  The first by the interval's method gives
                  $$1-sqrt5leq x<0$$ or $$xgeq1+sqrt5$$ or $$x<-10,$$ which with our domain gives the answer:
                  $$left[1-sqrt5,-frac{1}{3}right)cup[1+sqrt5,+infty).$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 23 at 14:36

























                  answered Jan 23 at 14:27









                  Michael RozenbergMichael Rozenberg

                  108k1895200




                  108k1895200























                      0












                      $begingroup$

                      Writing your inequality in the form $$frac{lnleft(6+frac{2}{x}right)}{ln(5)}+frac{lnleft(1+frac{x}{10}right)}{-ln(5)}le 1$$ we get the inequalities
                      $$6+frac{2}{x}>0$$ and $$1+frac{x}{10}>0$$ and
                      $$frac{6+frac{2}{x}}{1+frac{x}{10}}le 5$$ we get
                      $$1-sqrt{5}le x<-frac{1}{3}$$ or $$xgeq 1+sqrt{5}$$






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Writing your inequality in the form $$frac{lnleft(6+frac{2}{x}right)}{ln(5)}+frac{lnleft(1+frac{x}{10}right)}{-ln(5)}le 1$$ we get the inequalities
                        $$6+frac{2}{x}>0$$ and $$1+frac{x}{10}>0$$ and
                        $$frac{6+frac{2}{x}}{1+frac{x}{10}}le 5$$ we get
                        $$1-sqrt{5}le x<-frac{1}{3}$$ or $$xgeq 1+sqrt{5}$$






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Writing your inequality in the form $$frac{lnleft(6+frac{2}{x}right)}{ln(5)}+frac{lnleft(1+frac{x}{10}right)}{-ln(5)}le 1$$ we get the inequalities
                          $$6+frac{2}{x}>0$$ and $$1+frac{x}{10}>0$$ and
                          $$frac{6+frac{2}{x}}{1+frac{x}{10}}le 5$$ we get
                          $$1-sqrt{5}le x<-frac{1}{3}$$ or $$xgeq 1+sqrt{5}$$






                          share|cite|improve this answer











                          $endgroup$



                          Writing your inequality in the form $$frac{lnleft(6+frac{2}{x}right)}{ln(5)}+frac{lnleft(1+frac{x}{10}right)}{-ln(5)}le 1$$ we get the inequalities
                          $$6+frac{2}{x}>0$$ and $$1+frac{x}{10}>0$$ and
                          $$frac{6+frac{2}{x}}{1+frac{x}{10}}le 5$$ we get
                          $$1-sqrt{5}le x<-frac{1}{3}$$ or $$xgeq 1+sqrt{5}$$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 23 at 14:51









                          Bernard

                          122k741116




                          122k741116










                          answered Jan 23 at 14:41









                          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                          77.7k42866




                          77.7k42866






























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