How to solve in terms of one variable with an equation involving two variables?












2












$begingroup$


So I have been dealing with an extremely difficult equation: $x^{2a} = frac{x}{a}$ and am confused on how to solve it. I am wondering how to solve for $x $ in terms of $ a$ in the case of this question. If possible, give an explanation of the answer.






random-variables







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    So I have been dealing with an extremely difficult equation: $x^{2a} = frac{x}{a}$ and am confused on how to solve it. I am wondering how to solve for $x $ in terms of $ a$ in the case of this question. If possible, give an explanation of the answer.






    random-variables







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      So I have been dealing with an extremely difficult equation: $x^{2a} = frac{x}{a}$ and am confused on how to solve it. I am wondering how to solve for $x $ in terms of $ a$ in the case of this question. If possible, give an explanation of the answer.






      random-variables







      share|cite|improve this question











      $endgroup$




      So I have been dealing with an extremely difficult equation: $x^{2a} = frac{x}{a}$ and am confused on how to solve it. I am wondering how to solve for $x $ in terms of $ a$ in the case of this question. If possible, give an explanation of the answer.






      random-variables




      random-variables






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 4:56









      Martund

      1,575212




      1,575212










      asked Jan 6 at 4:41









      JamesJames

      264




      264






















          4 Answers
          4






          active

          oldest

          votes


















          1












          $begingroup$

          $$x^{2a}=x/a$$You can divide both sides by $x$ to get $$frac{x^{2a}}{x}=frac1a$$Use index rules to get $$x^{2a-1}=frac1a$$Then raise both sides to the power $frac1{2a-1}$ to get $$x=left(frac{1}aright)^{frac1{2a-1}}=a^{frac1{1-2a}}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How can Hello World's answer be simplified to your answer?
            $endgroup$
            – James
            Jan 6 at 5:03










          • $begingroup$
            $$e^{frac{ln a}{1-2a}}=(e^{ln a})^{frac1{1-2a}}=a^{frac1{1-2a}}$$Since $e^{ln a}=a$ by the fact that these are inverse functions of each other.
            $endgroup$
            – John Doe
            Jan 6 at 5:04





















          1












          $begingroup$

          I am assuming you want to solve:
          $$x^{2a} = frac{x}{a}implies 2aln(x)=ln(x)-ln(a)implies ln(x)=frac{ln(a)}{1-2a}$$
          $$implies x=e^{frac{ln(a)}{1-2a}} = left(e^{ln a}right)^{{1/(1-2a)}}=a^{frac{1}{1-2a}}.$$



          Another way would be to do:



          $$x^{2a} = frac{x}{a}implies x^{2a-1}=a^{-1}implies x=a^{frac{1}{1-2a}}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Which can be simplified to $a^{frac1{1-2a}}$. :)
            $endgroup$
            – John Doe
            Jan 6 at 4:59










          • $begingroup$
            How can this be simplified to your answer John Doe?
            $endgroup$
            – James
            Jan 6 at 5:01






          • 2




            $begingroup$
            My guess is: $$e^{ab} = (e^a)^b$$ Take $a$ to be the expression with the natural log here
            $endgroup$
            – Eevee Trainer
            Jan 6 at 5:03










          • $begingroup$
            Hello World, is there not a difference between a and e though?
            $endgroup$
            – James
            Jan 6 at 5:04










          • $begingroup$
            Also, in the last step of the second way to complete the equation, how can x^(2a-1)=a^-1 become x=a^(1/(1-2a))?
            $endgroup$
            – James
            Jan 6 at 5:07





















          0












          $begingroup$

          We have,
          $$x^{2a}=frac{x}{a}$$
          We see that $x=0$ is always a solution.



          Now, if $aneqfrac{1}{2}$, then,
          $$Longrightarrow x^{2a-1}=frac{1}{a}$$
          $$Longrightarrow x=a^{frac{1}{1-2a}}$$
          where both positive and negative solutions will come if $2a-1$ is even integer.



          For $a=frac{1}{2}$, $x=0$ is the only solution.



          Hope it is helpful:)






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry, I forgot to mention that x>0 and a>1.
            $endgroup$
            – James
            Jan 6 at 5:06










          • $begingroup$
            Ok then, ignore that for $a=frac{1}{2}$ and consider only positive solutions for $x$.
            $endgroup$
            – Martund
            Jan 6 at 5:08










          • $begingroup$
            How is the second to last step translated into the result?
            $endgroup$
            – James
            Jan 6 at 5:10










          • $begingroup$
            Take power $frac{1}{2a-1}$ on both sides, and apply $(m^p)^q=m^{pq}$ on both sides.
            $endgroup$
            – Martund
            Jan 6 at 5:14










          • $begingroup$
            So then on the side of x, the power of x would be 1?
            $endgroup$
            – James
            Jan 6 at 5:15



















          0












          $begingroup$

          This may be the way to proceed:
          Hope this helps.
          .






          share|cite|improve this answer









          $endgroup$













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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            $$x^{2a}=x/a$$You can divide both sides by $x$ to get $$frac{x^{2a}}{x}=frac1a$$Use index rules to get $$x^{2a-1}=frac1a$$Then raise both sides to the power $frac1{2a-1}$ to get $$x=left(frac{1}aright)^{frac1{2a-1}}=a^{frac1{1-2a}}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How can Hello World's answer be simplified to your answer?
              $endgroup$
              – James
              Jan 6 at 5:03










            • $begingroup$
              $$e^{frac{ln a}{1-2a}}=(e^{ln a})^{frac1{1-2a}}=a^{frac1{1-2a}}$$Since $e^{ln a}=a$ by the fact that these are inverse functions of each other.
              $endgroup$
              – John Doe
              Jan 6 at 5:04


















            1












            $begingroup$

            $$x^{2a}=x/a$$You can divide both sides by $x$ to get $$frac{x^{2a}}{x}=frac1a$$Use index rules to get $$x^{2a-1}=frac1a$$Then raise both sides to the power $frac1{2a-1}$ to get $$x=left(frac{1}aright)^{frac1{2a-1}}=a^{frac1{1-2a}}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How can Hello World's answer be simplified to your answer?
              $endgroup$
              – James
              Jan 6 at 5:03










            • $begingroup$
              $$e^{frac{ln a}{1-2a}}=(e^{ln a})^{frac1{1-2a}}=a^{frac1{1-2a}}$$Since $e^{ln a}=a$ by the fact that these are inverse functions of each other.
              $endgroup$
              – John Doe
              Jan 6 at 5:04
















            1












            1








            1





            $begingroup$

            $$x^{2a}=x/a$$You can divide both sides by $x$ to get $$frac{x^{2a}}{x}=frac1a$$Use index rules to get $$x^{2a-1}=frac1a$$Then raise both sides to the power $frac1{2a-1}$ to get $$x=left(frac{1}aright)^{frac1{2a-1}}=a^{frac1{1-2a}}$$






            share|cite|improve this answer









            $endgroup$



            $$x^{2a}=x/a$$You can divide both sides by $x$ to get $$frac{x^{2a}}{x}=frac1a$$Use index rules to get $$x^{2a-1}=frac1a$$Then raise both sides to the power $frac1{2a-1}$ to get $$x=left(frac{1}aright)^{frac1{2a-1}}=a^{frac1{1-2a}}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 4:59









            John DoeJohn Doe

            11.1k11238




            11.1k11238












            • $begingroup$
              How can Hello World's answer be simplified to your answer?
              $endgroup$
              – James
              Jan 6 at 5:03










            • $begingroup$
              $$e^{frac{ln a}{1-2a}}=(e^{ln a})^{frac1{1-2a}}=a^{frac1{1-2a}}$$Since $e^{ln a}=a$ by the fact that these are inverse functions of each other.
              $endgroup$
              – John Doe
              Jan 6 at 5:04




















            • $begingroup$
              How can Hello World's answer be simplified to your answer?
              $endgroup$
              – James
              Jan 6 at 5:03










            • $begingroup$
              $$e^{frac{ln a}{1-2a}}=(e^{ln a})^{frac1{1-2a}}=a^{frac1{1-2a}}$$Since $e^{ln a}=a$ by the fact that these are inverse functions of each other.
              $endgroup$
              – John Doe
              Jan 6 at 5:04


















            $begingroup$
            How can Hello World's answer be simplified to your answer?
            $endgroup$
            – James
            Jan 6 at 5:03




            $begingroup$
            How can Hello World's answer be simplified to your answer?
            $endgroup$
            – James
            Jan 6 at 5:03












            $begingroup$
            $$e^{frac{ln a}{1-2a}}=(e^{ln a})^{frac1{1-2a}}=a^{frac1{1-2a}}$$Since $e^{ln a}=a$ by the fact that these are inverse functions of each other.
            $endgroup$
            – John Doe
            Jan 6 at 5:04






            $begingroup$
            $$e^{frac{ln a}{1-2a}}=(e^{ln a})^{frac1{1-2a}}=a^{frac1{1-2a}}$$Since $e^{ln a}=a$ by the fact that these are inverse functions of each other.
            $endgroup$
            – John Doe
            Jan 6 at 5:04













            1












            $begingroup$

            I am assuming you want to solve:
            $$x^{2a} = frac{x}{a}implies 2aln(x)=ln(x)-ln(a)implies ln(x)=frac{ln(a)}{1-2a}$$
            $$implies x=e^{frac{ln(a)}{1-2a}} = left(e^{ln a}right)^{{1/(1-2a)}}=a^{frac{1}{1-2a}}.$$



            Another way would be to do:



            $$x^{2a} = frac{x}{a}implies x^{2a-1}=a^{-1}implies x=a^{frac{1}{1-2a}}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Which can be simplified to $a^{frac1{1-2a}}$. :)
              $endgroup$
              – John Doe
              Jan 6 at 4:59










            • $begingroup$
              How can this be simplified to your answer John Doe?
              $endgroup$
              – James
              Jan 6 at 5:01






            • 2




              $begingroup$
              My guess is: $$e^{ab} = (e^a)^b$$ Take $a$ to be the expression with the natural log here
              $endgroup$
              – Eevee Trainer
              Jan 6 at 5:03










            • $begingroup$
              Hello World, is there not a difference between a and e though?
              $endgroup$
              – James
              Jan 6 at 5:04










            • $begingroup$
              Also, in the last step of the second way to complete the equation, how can x^(2a-1)=a^-1 become x=a^(1/(1-2a))?
              $endgroup$
              – James
              Jan 6 at 5:07


















            1












            $begingroup$

            I am assuming you want to solve:
            $$x^{2a} = frac{x}{a}implies 2aln(x)=ln(x)-ln(a)implies ln(x)=frac{ln(a)}{1-2a}$$
            $$implies x=e^{frac{ln(a)}{1-2a}} = left(e^{ln a}right)^{{1/(1-2a)}}=a^{frac{1}{1-2a}}.$$



            Another way would be to do:



            $$x^{2a} = frac{x}{a}implies x^{2a-1}=a^{-1}implies x=a^{frac{1}{1-2a}}.$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Which can be simplified to $a^{frac1{1-2a}}$. :)
              $endgroup$
              – John Doe
              Jan 6 at 4:59










            • $begingroup$
              How can this be simplified to your answer John Doe?
              $endgroup$
              – James
              Jan 6 at 5:01






            • 2




              $begingroup$
              My guess is: $$e^{ab} = (e^a)^b$$ Take $a$ to be the expression with the natural log here
              $endgroup$
              – Eevee Trainer
              Jan 6 at 5:03










            • $begingroup$
              Hello World, is there not a difference between a and e though?
              $endgroup$
              – James
              Jan 6 at 5:04










            • $begingroup$
              Also, in the last step of the second way to complete the equation, how can x^(2a-1)=a^-1 become x=a^(1/(1-2a))?
              $endgroup$
              – James
              Jan 6 at 5:07
















            1












            1








            1





            $begingroup$

            I am assuming you want to solve:
            $$x^{2a} = frac{x}{a}implies 2aln(x)=ln(x)-ln(a)implies ln(x)=frac{ln(a)}{1-2a}$$
            $$implies x=e^{frac{ln(a)}{1-2a}} = left(e^{ln a}right)^{{1/(1-2a)}}=a^{frac{1}{1-2a}}.$$



            Another way would be to do:



            $$x^{2a} = frac{x}{a}implies x^{2a-1}=a^{-1}implies x=a^{frac{1}{1-2a}}.$$






            share|cite|improve this answer











            $endgroup$



            I am assuming you want to solve:
            $$x^{2a} = frac{x}{a}implies 2aln(x)=ln(x)-ln(a)implies ln(x)=frac{ln(a)}{1-2a}$$
            $$implies x=e^{frac{ln(a)}{1-2a}} = left(e^{ln a}right)^{{1/(1-2a)}}=a^{frac{1}{1-2a}}.$$



            Another way would be to do:



            $$x^{2a} = frac{x}{a}implies x^{2a-1}=a^{-1}implies x=a^{frac{1}{1-2a}}.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 6 at 5:05

























            answered Jan 6 at 4:57









            Hello_WorldHello_World

            4,12621731




            4,12621731












            • $begingroup$
              Which can be simplified to $a^{frac1{1-2a}}$. :)
              $endgroup$
              – John Doe
              Jan 6 at 4:59










            • $begingroup$
              How can this be simplified to your answer John Doe?
              $endgroup$
              – James
              Jan 6 at 5:01






            • 2




              $begingroup$
              My guess is: $$e^{ab} = (e^a)^b$$ Take $a$ to be the expression with the natural log here
              $endgroup$
              – Eevee Trainer
              Jan 6 at 5:03










            • $begingroup$
              Hello World, is there not a difference between a and e though?
              $endgroup$
              – James
              Jan 6 at 5:04










            • $begingroup$
              Also, in the last step of the second way to complete the equation, how can x^(2a-1)=a^-1 become x=a^(1/(1-2a))?
              $endgroup$
              – James
              Jan 6 at 5:07




















            • $begingroup$
              Which can be simplified to $a^{frac1{1-2a}}$. :)
              $endgroup$
              – John Doe
              Jan 6 at 4:59










            • $begingroup$
              How can this be simplified to your answer John Doe?
              $endgroup$
              – James
              Jan 6 at 5:01






            • 2




              $begingroup$
              My guess is: $$e^{ab} = (e^a)^b$$ Take $a$ to be the expression with the natural log here
              $endgroup$
              – Eevee Trainer
              Jan 6 at 5:03










            • $begingroup$
              Hello World, is there not a difference between a and e though?
              $endgroup$
              – James
              Jan 6 at 5:04










            • $begingroup$
              Also, in the last step of the second way to complete the equation, how can x^(2a-1)=a^-1 become x=a^(1/(1-2a))?
              $endgroup$
              – James
              Jan 6 at 5:07


















            $begingroup$
            Which can be simplified to $a^{frac1{1-2a}}$. :)
            $endgroup$
            – John Doe
            Jan 6 at 4:59




            $begingroup$
            Which can be simplified to $a^{frac1{1-2a}}$. :)
            $endgroup$
            – John Doe
            Jan 6 at 4:59












            $begingroup$
            How can this be simplified to your answer John Doe?
            $endgroup$
            – James
            Jan 6 at 5:01




            $begingroup$
            How can this be simplified to your answer John Doe?
            $endgroup$
            – James
            Jan 6 at 5:01




            2




            2




            $begingroup$
            My guess is: $$e^{ab} = (e^a)^b$$ Take $a$ to be the expression with the natural log here
            $endgroup$
            – Eevee Trainer
            Jan 6 at 5:03




            $begingroup$
            My guess is: $$e^{ab} = (e^a)^b$$ Take $a$ to be the expression with the natural log here
            $endgroup$
            – Eevee Trainer
            Jan 6 at 5:03












            $begingroup$
            Hello World, is there not a difference between a and e though?
            $endgroup$
            – James
            Jan 6 at 5:04




            $begingroup$
            Hello World, is there not a difference between a and e though?
            $endgroup$
            – James
            Jan 6 at 5:04












            $begingroup$
            Also, in the last step of the second way to complete the equation, how can x^(2a-1)=a^-1 become x=a^(1/(1-2a))?
            $endgroup$
            – James
            Jan 6 at 5:07






            $begingroup$
            Also, in the last step of the second way to complete the equation, how can x^(2a-1)=a^-1 become x=a^(1/(1-2a))?
            $endgroup$
            – James
            Jan 6 at 5:07













            0












            $begingroup$

            We have,
            $$x^{2a}=frac{x}{a}$$
            We see that $x=0$ is always a solution.



            Now, if $aneqfrac{1}{2}$, then,
            $$Longrightarrow x^{2a-1}=frac{1}{a}$$
            $$Longrightarrow x=a^{frac{1}{1-2a}}$$
            where both positive and negative solutions will come if $2a-1$ is even integer.



            For $a=frac{1}{2}$, $x=0$ is the only solution.



            Hope it is helpful:)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Sorry, I forgot to mention that x>0 and a>1.
              $endgroup$
              – James
              Jan 6 at 5:06










            • $begingroup$
              Ok then, ignore that for $a=frac{1}{2}$ and consider only positive solutions for $x$.
              $endgroup$
              – Martund
              Jan 6 at 5:08










            • $begingroup$
              How is the second to last step translated into the result?
              $endgroup$
              – James
              Jan 6 at 5:10










            • $begingroup$
              Take power $frac{1}{2a-1}$ on both sides, and apply $(m^p)^q=m^{pq}$ on both sides.
              $endgroup$
              – Martund
              Jan 6 at 5:14










            • $begingroup$
              So then on the side of x, the power of x would be 1?
              $endgroup$
              – James
              Jan 6 at 5:15
















            0












            $begingroup$

            We have,
            $$x^{2a}=frac{x}{a}$$
            We see that $x=0$ is always a solution.



            Now, if $aneqfrac{1}{2}$, then,
            $$Longrightarrow x^{2a-1}=frac{1}{a}$$
            $$Longrightarrow x=a^{frac{1}{1-2a}}$$
            where both positive and negative solutions will come if $2a-1$ is even integer.



            For $a=frac{1}{2}$, $x=0$ is the only solution.



            Hope it is helpful:)






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Sorry, I forgot to mention that x>0 and a>1.
              $endgroup$
              – James
              Jan 6 at 5:06










            • $begingroup$
              Ok then, ignore that for $a=frac{1}{2}$ and consider only positive solutions for $x$.
              $endgroup$
              – Martund
              Jan 6 at 5:08










            • $begingroup$
              How is the second to last step translated into the result?
              $endgroup$
              – James
              Jan 6 at 5:10










            • $begingroup$
              Take power $frac{1}{2a-1}$ on both sides, and apply $(m^p)^q=m^{pq}$ on both sides.
              $endgroup$
              – Martund
              Jan 6 at 5:14










            • $begingroup$
              So then on the side of x, the power of x would be 1?
              $endgroup$
              – James
              Jan 6 at 5:15














            0












            0








            0





            $begingroup$

            We have,
            $$x^{2a}=frac{x}{a}$$
            We see that $x=0$ is always a solution.



            Now, if $aneqfrac{1}{2}$, then,
            $$Longrightarrow x^{2a-1}=frac{1}{a}$$
            $$Longrightarrow x=a^{frac{1}{1-2a}}$$
            where both positive and negative solutions will come if $2a-1$ is even integer.



            For $a=frac{1}{2}$, $x=0$ is the only solution.



            Hope it is helpful:)






            share|cite|improve this answer









            $endgroup$



            We have,
            $$x^{2a}=frac{x}{a}$$
            We see that $x=0$ is always a solution.



            Now, if $aneqfrac{1}{2}$, then,
            $$Longrightarrow x^{2a-1}=frac{1}{a}$$
            $$Longrightarrow x=a^{frac{1}{1-2a}}$$
            where both positive and negative solutions will come if $2a-1$ is even integer.



            For $a=frac{1}{2}$, $x=0$ is the only solution.



            Hope it is helpful:)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 5:04









            MartundMartund

            1,575212




            1,575212












            • $begingroup$
              Sorry, I forgot to mention that x>0 and a>1.
              $endgroup$
              – James
              Jan 6 at 5:06










            • $begingroup$
              Ok then, ignore that for $a=frac{1}{2}$ and consider only positive solutions for $x$.
              $endgroup$
              – Martund
              Jan 6 at 5:08










            • $begingroup$
              How is the second to last step translated into the result?
              $endgroup$
              – James
              Jan 6 at 5:10










            • $begingroup$
              Take power $frac{1}{2a-1}$ on both sides, and apply $(m^p)^q=m^{pq}$ on both sides.
              $endgroup$
              – Martund
              Jan 6 at 5:14










            • $begingroup$
              So then on the side of x, the power of x would be 1?
              $endgroup$
              – James
              Jan 6 at 5:15


















            • $begingroup$
              Sorry, I forgot to mention that x>0 and a>1.
              $endgroup$
              – James
              Jan 6 at 5:06










            • $begingroup$
              Ok then, ignore that for $a=frac{1}{2}$ and consider only positive solutions for $x$.
              $endgroup$
              – Martund
              Jan 6 at 5:08










            • $begingroup$
              How is the second to last step translated into the result?
              $endgroup$
              – James
              Jan 6 at 5:10










            • $begingroup$
              Take power $frac{1}{2a-1}$ on both sides, and apply $(m^p)^q=m^{pq}$ on both sides.
              $endgroup$
              – Martund
              Jan 6 at 5:14










            • $begingroup$
              So then on the side of x, the power of x would be 1?
              $endgroup$
              – James
              Jan 6 at 5:15
















            $begingroup$
            Sorry, I forgot to mention that x>0 and a>1.
            $endgroup$
            – James
            Jan 6 at 5:06




            $begingroup$
            Sorry, I forgot to mention that x>0 and a>1.
            $endgroup$
            – James
            Jan 6 at 5:06












            $begingroup$
            Ok then, ignore that for $a=frac{1}{2}$ and consider only positive solutions for $x$.
            $endgroup$
            – Martund
            Jan 6 at 5:08




            $begingroup$
            Ok then, ignore that for $a=frac{1}{2}$ and consider only positive solutions for $x$.
            $endgroup$
            – Martund
            Jan 6 at 5:08












            $begingroup$
            How is the second to last step translated into the result?
            $endgroup$
            – James
            Jan 6 at 5:10




            $begingroup$
            How is the second to last step translated into the result?
            $endgroup$
            – James
            Jan 6 at 5:10












            $begingroup$
            Take power $frac{1}{2a-1}$ on both sides, and apply $(m^p)^q=m^{pq}$ on both sides.
            $endgroup$
            – Martund
            Jan 6 at 5:14




            $begingroup$
            Take power $frac{1}{2a-1}$ on both sides, and apply $(m^p)^q=m^{pq}$ on both sides.
            $endgroup$
            – Martund
            Jan 6 at 5:14












            $begingroup$
            So then on the side of x, the power of x would be 1?
            $endgroup$
            – James
            Jan 6 at 5:15




            $begingroup$
            So then on the side of x, the power of x would be 1?
            $endgroup$
            – James
            Jan 6 at 5:15











            0












            $begingroup$

            This may be the way to proceed:
            Hope this helps.
            .






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              This may be the way to proceed:
              Hope this helps.
              .






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                This may be the way to proceed:
                Hope this helps.
                .






                share|cite|improve this answer









                $endgroup$



                This may be the way to proceed:
                Hope this helps.
                .







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 at 5:39









                user629353user629353

                1187




                1187






























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