Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$












0












$begingroup$



Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$




I equated their exponents,



That gave me $log_5 x = frac{8}{3}$



But the answer given in my book is $1$.



Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.



When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?



Any help would be appreciated.






functions logarithms







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$endgroup$












  • $begingroup$
    What does your calculator show if you plot the range $x=50..100?$
    $endgroup$
    – gammatester
    Dec 6 '18 at 16:36


















0












$begingroup$



Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$




I equated their exponents,



That gave me $log_5 x = frac{8}{3}$



But the answer given in my book is $1$.



Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.



When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?



Any help would be appreciated.






functions logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does your calculator show if you plot the range $x=50..100?$
    $endgroup$
    – gammatester
    Dec 6 '18 at 16:36
















0












0








0





$begingroup$



Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$




I equated their exponents,



That gave me $log_5 x = frac{8}{3}$



But the answer given in my book is $1$.



Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.



When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?



Any help would be appreciated.






functions logarithms







share|cite|improve this question











$endgroup$





Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$




I equated their exponents,



That gave me $log_5 x = frac{8}{3}$



But the answer given in my book is $1$.



Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.



When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?



Any help would be appreciated.






functions logarithms




functions logarithms






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share|cite|improve this question













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edited Dec 8 '18 at 11:48









N. F. Taussig

44k93356




44k93356










asked Dec 6 '18 at 16:05









Piano LandPiano Land

379115




379115












  • $begingroup$
    What does your calculator show if you plot the range $x=50..100?$
    $endgroup$
    – gammatester
    Dec 6 '18 at 16:36




















  • $begingroup$
    What does your calculator show if you plot the range $x=50..100?$
    $endgroup$
    – gammatester
    Dec 6 '18 at 16:36


















$begingroup$
What does your calculator show if you plot the range $x=50..100?$
$endgroup$
– gammatester
Dec 6 '18 at 16:36






$begingroup$
What does your calculator show if you plot the range $x=50..100?$
$endgroup$
– gammatester
Dec 6 '18 at 16:36












5 Answers
5






active

oldest

votes


















1












$begingroup$

You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
$$
(log_5(x^2)+log_5x-12)log_5x=-4log_5x
$$
that becomes, setting $y=log_5x$,
$$
(3y-8)y=0
$$
so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.



On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
$$
(y^2+y-8)y=0
$$
with solutions
$$
y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I suppose you were using
    $$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
    but it is only true when $x>0$ and $xne 1$.



    Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
      $endgroup$
      – Piano Land
      Dec 6 '18 at 16:25










    • $begingroup$
      I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
      $endgroup$
      – Eclipse Sun
      Dec 6 '18 at 16:28










    • $begingroup$
      Perhaps just because $x=5^{8/3}$ is too large.
      $endgroup$
      – Eclipse Sun
      Dec 6 '18 at 16:38



















    1












    $begingroup$

    We need



    $$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$



    that is



    $$x=5^frac83$$



    the other solution $x=1$ is obtained by inspection from the original equation.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @KM101 Opssss...thanks I fix!
      $endgroup$
      – gimusi
      Dec 6 '18 at 16:26










    • $begingroup$
      No problem! :-)
      $endgroup$
      – KM101
      Dec 6 '18 at 16:36





















    1












    $begingroup$

    If the book claims, that $1$ is the only real solution, it is wrong. Your value
    $$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      What you did (probably) was the following:



      $$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$



      By doing so, you removed the possibility of $x = 1$.



      As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.




      Your other solution is valid. Let $x = 5^{frac{8}{3}}$.



      $$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$



      On both sides, you get



      $$big(5^{frac{8}{3}}big)^{-4}$$






      share|cite|improve this answer











      $endgroup$













        Your Answer





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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
        $$
        (log_5(x^2)+log_5x-12)log_5x=-4log_5x
        $$
        that becomes, setting $y=log_5x$,
        $$
        (3y-8)y=0
        $$
        so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.



        On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
        $$
        (y^2+y-8)y=0
        $$
        with solutions
        $$
        y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
        $$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
          $$
          (log_5(x^2)+log_5x-12)log_5x=-4log_5x
          $$
          that becomes, setting $y=log_5x$,
          $$
          (3y-8)y=0
          $$
          so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.



          On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
          $$
          (y^2+y-8)y=0
          $$
          with solutions
          $$
          y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
          $$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
            $$
            (log_5(x^2)+log_5x-12)log_5x=-4log_5x
            $$
            that becomes, setting $y=log_5x$,
            $$
            (3y-8)y=0
            $$
            so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.



            On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
            $$
            (y^2+y-8)y=0
            $$
            with solutions
            $$
            y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
            $$






            share|cite|improve this answer









            $endgroup$



            You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
            $$
            (log_5(x^2)+log_5x-12)log_5x=-4log_5x
            $$
            that becomes, setting $y=log_5x$,
            $$
            (3y-8)y=0
            $$
            so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.



            On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
            $$
            (y^2+y-8)y=0
            $$
            with solutions
            $$
            y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 6 '18 at 16:49









            egregegreg

            180k1485202




            180k1485202























                1












                $begingroup$

                I suppose you were using
                $$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
                but it is only true when $x>0$ and $xne 1$.



                Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                  $endgroup$
                  – Piano Land
                  Dec 6 '18 at 16:25










                • $begingroup$
                  I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:28










                • $begingroup$
                  Perhaps just because $x=5^{8/3}$ is too large.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:38
















                1












                $begingroup$

                I suppose you were using
                $$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
                but it is only true when $x>0$ and $xne 1$.



                Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                  $endgroup$
                  – Piano Land
                  Dec 6 '18 at 16:25










                • $begingroup$
                  I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:28










                • $begingroup$
                  Perhaps just because $x=5^{8/3}$ is too large.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:38














                1












                1








                1





                $begingroup$

                I suppose you were using
                $$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
                but it is only true when $x>0$ and $xne 1$.



                Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.






                share|cite|improve this answer









                $endgroup$



                I suppose you were using
                $$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
                but it is only true when $x>0$ and $xne 1$.



                Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 16:11









                Eclipse SunEclipse Sun

                6,9941437




                6,9941437












                • $begingroup$
                  But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                  $endgroup$
                  – Piano Land
                  Dec 6 '18 at 16:25










                • $begingroup$
                  I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:28










                • $begingroup$
                  Perhaps just because $x=5^{8/3}$ is too large.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:38


















                • $begingroup$
                  But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                  $endgroup$
                  – Piano Land
                  Dec 6 '18 at 16:25










                • $begingroup$
                  I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:28










                • $begingroup$
                  Perhaps just because $x=5^{8/3}$ is too large.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:38
















                $begingroup$
                But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                $endgroup$
                – Piano Land
                Dec 6 '18 at 16:25




                $begingroup$
                But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                $endgroup$
                – Piano Land
                Dec 6 '18 at 16:25












                $begingroup$
                I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                $endgroup$
                – Eclipse Sun
                Dec 6 '18 at 16:28




                $begingroup$
                I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                $endgroup$
                – Eclipse Sun
                Dec 6 '18 at 16:28












                $begingroup$
                Perhaps just because $x=5^{8/3}$ is too large.
                $endgroup$
                – Eclipse Sun
                Dec 6 '18 at 16:38




                $begingroup$
                Perhaps just because $x=5^{8/3}$ is too large.
                $endgroup$
                – Eclipse Sun
                Dec 6 '18 at 16:38











                1












                $begingroup$

                We need



                $$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$



                that is



                $$x=5^frac83$$



                the other solution $x=1$ is obtained by inspection from the original equation.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  @KM101 Opssss...thanks I fix!
                  $endgroup$
                  – gimusi
                  Dec 6 '18 at 16:26










                • $begingroup$
                  No problem! :-)
                  $endgroup$
                  – KM101
                  Dec 6 '18 at 16:36


















                1












                $begingroup$

                We need



                $$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$



                that is



                $$x=5^frac83$$



                the other solution $x=1$ is obtained by inspection from the original equation.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  @KM101 Opssss...thanks I fix!
                  $endgroup$
                  – gimusi
                  Dec 6 '18 at 16:26










                • $begingroup$
                  No problem! :-)
                  $endgroup$
                  – KM101
                  Dec 6 '18 at 16:36
















                1












                1








                1





                $begingroup$

                We need



                $$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$



                that is



                $$x=5^frac83$$



                the other solution $x=1$ is obtained by inspection from the original equation.






                share|cite|improve this answer











                $endgroup$



                We need



                $$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$



                that is



                $$x=5^frac83$$



                the other solution $x=1$ is obtained by inspection from the original equation.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 6 '18 at 16:27

























                answered Dec 6 '18 at 16:24









                gimusigimusi

                92.8k84494




                92.8k84494












                • $begingroup$
                  @KM101 Opssss...thanks I fix!
                  $endgroup$
                  – gimusi
                  Dec 6 '18 at 16:26










                • $begingroup$
                  No problem! :-)
                  $endgroup$
                  – KM101
                  Dec 6 '18 at 16:36




















                • $begingroup$
                  @KM101 Opssss...thanks I fix!
                  $endgroup$
                  – gimusi
                  Dec 6 '18 at 16:26










                • $begingroup$
                  No problem! :-)
                  $endgroup$
                  – KM101
                  Dec 6 '18 at 16:36


















                $begingroup$
                @KM101 Opssss...thanks I fix!
                $endgroup$
                – gimusi
                Dec 6 '18 at 16:26




                $begingroup$
                @KM101 Opssss...thanks I fix!
                $endgroup$
                – gimusi
                Dec 6 '18 at 16:26












                $begingroup$
                No problem! :-)
                $endgroup$
                – KM101
                Dec 6 '18 at 16:36






                $begingroup$
                No problem! :-)
                $endgroup$
                – KM101
                Dec 6 '18 at 16:36













                1












                $begingroup$

                If the book claims, that $1$ is the only real solution, it is wrong. Your value
                $$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  If the book claims, that $1$ is the only real solution, it is wrong. Your value
                  $$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    If the book claims, that $1$ is the only real solution, it is wrong. Your value
                    $$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.






                    share|cite|improve this answer











                    $endgroup$



                    If the book claims, that $1$ is the only real solution, it is wrong. Your value
                    $$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 6 '18 at 16:28

























                    answered Dec 6 '18 at 16:22









                    gammatestergammatester

                    16.7k21632




                    16.7k21632























                        1












                        $begingroup$

                        What you did (probably) was the following:



                        $$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$



                        By doing so, you removed the possibility of $x = 1$.



                        As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.




                        Your other solution is valid. Let $x = 5^{frac{8}{3}}$.



                        $$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$



                        On both sides, you get



                        $$big(5^{frac{8}{3}}big)^{-4}$$






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          What you did (probably) was the following:



                          $$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$



                          By doing so, you removed the possibility of $x = 1$.



                          As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.




                          Your other solution is valid. Let $x = 5^{frac{8}{3}}$.



                          $$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$



                          On both sides, you get



                          $$big(5^{frac{8}{3}}big)^{-4}$$






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            What you did (probably) was the following:



                            $$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$



                            By doing so, you removed the possibility of $x = 1$.



                            As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.




                            Your other solution is valid. Let $x = 5^{frac{8}{3}}$.



                            $$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$



                            On both sides, you get



                            $$big(5^{frac{8}{3}}big)^{-4}$$






                            share|cite|improve this answer











                            $endgroup$



                            What you did (probably) was the following:



                            $$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$



                            By doing so, you removed the possibility of $x = 1$.



                            As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.




                            Your other solution is valid. Let $x = 5^{frac{8}{3}}$.



                            $$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$



                            On both sides, you get



                            $$big(5^{frac{8}{3}}big)^{-4}$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 6 '18 at 16:36

























                            answered Dec 6 '18 at 16:22









                            KM101KM101

                            5,9281524




                            5,9281524






























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