$log_{0.1}(x^4) - 4 geq 0$ - Solution verification












0












$begingroup$


My question: are these steps ok?

To be more precise, is the step where I take fourth root ok?
$$log_{0.1}(x^4) - 4 geq 0$$
$$iff$$
$$log_{0.1}(x^4) geq 4$$
$$iff$$
$$log_{0.1}(x^4) geq log_{0.1}left(frac{1}{10000}right)$$
$$iff$$
$$x^4 leq frac{1}{10000}$$
$$iff$$
$$|x|leq frac{1}{10}$$
This is the solution without $x=0$.






proof-verification logarithms absolute-value







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$endgroup$












  • $begingroup$
    Sounds right to me...
    $endgroup$
    – Mostafa Ayaz
    Jan 12 at 8:50
















0












$begingroup$


My question: are these steps ok?

To be more precise, is the step where I take fourth root ok?
$$log_{0.1}(x^4) - 4 geq 0$$
$$iff$$
$$log_{0.1}(x^4) geq 4$$
$$iff$$
$$log_{0.1}(x^4) geq log_{0.1}left(frac{1}{10000}right)$$
$$iff$$
$$x^4 leq frac{1}{10000}$$
$$iff$$
$$|x|leq frac{1}{10}$$
This is the solution without $x=0$.






proof-verification logarithms absolute-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    Sounds right to me...
    $endgroup$
    – Mostafa Ayaz
    Jan 12 at 8:50














0












0








0





$begingroup$


My question: are these steps ok?

To be more precise, is the step where I take fourth root ok?
$$log_{0.1}(x^4) - 4 geq 0$$
$$iff$$
$$log_{0.1}(x^4) geq 4$$
$$iff$$
$$log_{0.1}(x^4) geq log_{0.1}left(frac{1}{10000}right)$$
$$iff$$
$$x^4 leq frac{1}{10000}$$
$$iff$$
$$|x|leq frac{1}{10}$$
This is the solution without $x=0$.






proof-verification logarithms absolute-value







share|cite|improve this question











$endgroup$




My question: are these steps ok?

To be more precise, is the step where I take fourth root ok?
$$log_{0.1}(x^4) - 4 geq 0$$
$$iff$$
$$log_{0.1}(x^4) geq 4$$
$$iff$$
$$log_{0.1}(x^4) geq log_{0.1}left(frac{1}{10000}right)$$
$$iff$$
$$x^4 leq frac{1}{10000}$$
$$iff$$
$$|x|leq frac{1}{10}$$
This is the solution without $x=0$.






proof-verification logarithms absolute-value




proof-verification logarithms absolute-value






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share|cite|improve this question













share|cite|improve this question




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edited Jan 17 at 12:53









Robert Z

102k1072145




102k1072145










asked Jan 12 at 8:47









josfjosf

296317




296317












  • $begingroup$
    Sounds right to me...
    $endgroup$
    – Mostafa Ayaz
    Jan 12 at 8:50


















  • $begingroup$
    Sounds right to me...
    $endgroup$
    – Mostafa Ayaz
    Jan 12 at 8:50
















$begingroup$
Sounds right to me...
$endgroup$
– Mostafa Ayaz
Jan 12 at 8:50




$begingroup$
Sounds right to me...
$endgroup$
– Mostafa Ayaz
Jan 12 at 8:50










3 Answers
3






active

oldest

votes


















1












$begingroup$

It is correct, but it would be easier to take into account that $log_{0.1}(x^4)=4log_{0.1}(x)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hmm...only true if $x>0$ of course.
    $endgroup$
    – drhab
    Jan 12 at 9:00










  • $begingroup$
    $vert xvert$ would be correct.
    $endgroup$
    – KM101
    Jan 12 at 9:03












  • $begingroup$
    @drhab RIght, but it is clear from the question that the OP is aware of that.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 9:05



















1












$begingroup$

It’s completely fine, and you also correctly used the absolute value when taking the fourth root, but you can solve the inequality more quickly by noting:



$$log_a b^c = clog_a vert bvert tag{1}$$



$$log_{a^b} c = frac{1}{b}log_a c tag{2}$$



So, using $0.1 = 10^{-1}$, you can rewrite $log_{0.1} left(x^4right) geq 4$ as follows:



$$log_{0.1} left(x^4right) geq 4 iff -4log_{10} vert xvert geq 4 iff log_{10} vert xvert leq -1 iff vert xvert leq frac{1}{10}$$



Of course, as mentioned, your way is completely fine as well, but perhaps this way is a bit faster.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    For fun, an option :



    $z:= log_{0.1}(x^4) ge 4;$



    $x^4= (0.1)^z=(10^{-1})^z= (10)^{-z}$, i.e.



    $(10)^z=x^{-4} ge (10)^4,$



    finally $x le 1/(10)$.



    Your answer is fine as has been pointed out.






    share|cite|improve this answer









    $endgroup$














      Your Answer








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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      It is correct, but it would be easier to take into account that $log_{0.1}(x^4)=4log_{0.1}(x)$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Hmm...only true if $x>0$ of course.
        $endgroup$
        – drhab
        Jan 12 at 9:00










      • $begingroup$
        $vert xvert$ would be correct.
        $endgroup$
        – KM101
        Jan 12 at 9:03












      • $begingroup$
        @drhab RIght, but it is clear from the question that the OP is aware of that.
        $endgroup$
        – José Carlos Santos
        Jan 12 at 9:05
















      1












      $begingroup$

      It is correct, but it would be easier to take into account that $log_{0.1}(x^4)=4log_{0.1}(x)$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Hmm...only true if $x>0$ of course.
        $endgroup$
        – drhab
        Jan 12 at 9:00










      • $begingroup$
        $vert xvert$ would be correct.
        $endgroup$
        – KM101
        Jan 12 at 9:03












      • $begingroup$
        @drhab RIght, but it is clear from the question that the OP is aware of that.
        $endgroup$
        – José Carlos Santos
        Jan 12 at 9:05














      1












      1








      1





      $begingroup$

      It is correct, but it would be easier to take into account that $log_{0.1}(x^4)=4log_{0.1}(x)$.






      share|cite|improve this answer









      $endgroup$



      It is correct, but it would be easier to take into account that $log_{0.1}(x^4)=4log_{0.1}(x)$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 12 at 8:57









      José Carlos SantosJosé Carlos Santos

      174k23134243




      174k23134243












      • $begingroup$
        Hmm...only true if $x>0$ of course.
        $endgroup$
        – drhab
        Jan 12 at 9:00










      • $begingroup$
        $vert xvert$ would be correct.
        $endgroup$
        – KM101
        Jan 12 at 9:03












      • $begingroup$
        @drhab RIght, but it is clear from the question that the OP is aware of that.
        $endgroup$
        – José Carlos Santos
        Jan 12 at 9:05


















      • $begingroup$
        Hmm...only true if $x>0$ of course.
        $endgroup$
        – drhab
        Jan 12 at 9:00










      • $begingroup$
        $vert xvert$ would be correct.
        $endgroup$
        – KM101
        Jan 12 at 9:03












      • $begingroup$
        @drhab RIght, but it is clear from the question that the OP is aware of that.
        $endgroup$
        – José Carlos Santos
        Jan 12 at 9:05
















      $begingroup$
      Hmm...only true if $x>0$ of course.
      $endgroup$
      – drhab
      Jan 12 at 9:00




      $begingroup$
      Hmm...only true if $x>0$ of course.
      $endgroup$
      – drhab
      Jan 12 at 9:00












      $begingroup$
      $vert xvert$ would be correct.
      $endgroup$
      – KM101
      Jan 12 at 9:03






      $begingroup$
      $vert xvert$ would be correct.
      $endgroup$
      – KM101
      Jan 12 at 9:03














      $begingroup$
      @drhab RIght, but it is clear from the question that the OP is aware of that.
      $endgroup$
      – José Carlos Santos
      Jan 12 at 9:05




      $begingroup$
      @drhab RIght, but it is clear from the question that the OP is aware of that.
      $endgroup$
      – José Carlos Santos
      Jan 12 at 9:05











      1












      $begingroup$

      It’s completely fine, and you also correctly used the absolute value when taking the fourth root, but you can solve the inequality more quickly by noting:



      $$log_a b^c = clog_a vert bvert tag{1}$$



      $$log_{a^b} c = frac{1}{b}log_a c tag{2}$$



      So, using $0.1 = 10^{-1}$, you can rewrite $log_{0.1} left(x^4right) geq 4$ as follows:



      $$log_{0.1} left(x^4right) geq 4 iff -4log_{10} vert xvert geq 4 iff log_{10} vert xvert leq -1 iff vert xvert leq frac{1}{10}$$



      Of course, as mentioned, your way is completely fine as well, but perhaps this way is a bit faster.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        It’s completely fine, and you also correctly used the absolute value when taking the fourth root, but you can solve the inequality more quickly by noting:



        $$log_a b^c = clog_a vert bvert tag{1}$$



        $$log_{a^b} c = frac{1}{b}log_a c tag{2}$$



        So, using $0.1 = 10^{-1}$, you can rewrite $log_{0.1} left(x^4right) geq 4$ as follows:



        $$log_{0.1} left(x^4right) geq 4 iff -4log_{10} vert xvert geq 4 iff log_{10} vert xvert leq -1 iff vert xvert leq frac{1}{10}$$



        Of course, as mentioned, your way is completely fine as well, but perhaps this way is a bit faster.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          It’s completely fine, and you also correctly used the absolute value when taking the fourth root, but you can solve the inequality more quickly by noting:



          $$log_a b^c = clog_a vert bvert tag{1}$$



          $$log_{a^b} c = frac{1}{b}log_a c tag{2}$$



          So, using $0.1 = 10^{-1}$, you can rewrite $log_{0.1} left(x^4right) geq 4$ as follows:



          $$log_{0.1} left(x^4right) geq 4 iff -4log_{10} vert xvert geq 4 iff log_{10} vert xvert leq -1 iff vert xvert leq frac{1}{10}$$



          Of course, as mentioned, your way is completely fine as well, but perhaps this way is a bit faster.






          share|cite|improve this answer









          $endgroup$



          It’s completely fine, and you also correctly used the absolute value when taking the fourth root, but you can solve the inequality more quickly by noting:



          $$log_a b^c = clog_a vert bvert tag{1}$$



          $$log_{a^b} c = frac{1}{b}log_a c tag{2}$$



          So, using $0.1 = 10^{-1}$, you can rewrite $log_{0.1} left(x^4right) geq 4$ as follows:



          $$log_{0.1} left(x^4right) geq 4 iff -4log_{10} vert xvert geq 4 iff log_{10} vert xvert leq -1 iff vert xvert leq frac{1}{10}$$



          Of course, as mentioned, your way is completely fine as well, but perhaps this way is a bit faster.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 12 at 9:06









          KM101KM101

          6,0861525




          6,0861525























              0












              $begingroup$

              For fun, an option :



              $z:= log_{0.1}(x^4) ge 4;$



              $x^4= (0.1)^z=(10^{-1})^z= (10)^{-z}$, i.e.



              $(10)^z=x^{-4} ge (10)^4,$



              finally $x le 1/(10)$.



              Your answer is fine as has been pointed out.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                For fun, an option :



                $z:= log_{0.1}(x^4) ge 4;$



                $x^4= (0.1)^z=(10^{-1})^z= (10)^{-z}$, i.e.



                $(10)^z=x^{-4} ge (10)^4,$



                finally $x le 1/(10)$.



                Your answer is fine as has been pointed out.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  For fun, an option :



                  $z:= log_{0.1}(x^4) ge 4;$



                  $x^4= (0.1)^z=(10^{-1})^z= (10)^{-z}$, i.e.



                  $(10)^z=x^{-4} ge (10)^4,$



                  finally $x le 1/(10)$.



                  Your answer is fine as has been pointed out.






                  share|cite|improve this answer









                  $endgroup$



                  For fun, an option :



                  $z:= log_{0.1}(x^4) ge 4;$



                  $x^4= (0.1)^z=(10^{-1})^z= (10)^{-z}$, i.e.



                  $(10)^z=x^{-4} ge (10)^4,$



                  finally $x le 1/(10)$.



                  Your answer is fine as has been pointed out.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 12 at 10:50









                  Peter SzilasPeter Szilas

                  12k2822




                  12k2822






























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