Exponential/Logarithmic equation system












3












$begingroup$



Solve the following equation system over the real numbers
$$begin{cases}
x(1-log_{10}(5))=log_{10}(11-3^y)\
log_{10}(35-4^x)=ylog_{10}(9) \
end{cases}
$$





For the functions in the above relations to be well defined, I've determined
that $0<x<log_4(35)$ and $0<y<log_3(11)$. I also know that each of the equations(taken separately by itself) has exactly one solution because of their monotonicity (in each case one is strictly increasing and the other is strictly decreasing, so they can't have more than one intersection.)




By substituting $(1-log_{10}(5))$ with $log_{10}(2)$ in the first equation the unique solution for it is found to be $x=log_2(11-3^y).$


I've tried to proceed further but to no avail and come here looking for help; I think that there is no real solution to the system but haven't been able to prove it.






algebra-precalculus logarithms exponential-function systems-of-equations







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$endgroup$












  • $begingroup$
    Base of the logarithm $lg$?
    $endgroup$
    – Robert Z
    Feb 16 at 11:22










  • $begingroup$
    10 is the base of lg of course
    $endgroup$
    – Luca Pana
    Feb 16 at 11:22
















3












$begingroup$



Solve the following equation system over the real numbers
$$begin{cases}
x(1-log_{10}(5))=log_{10}(11-3^y)\
log_{10}(35-4^x)=ylog_{10}(9) \
end{cases}
$$





For the functions in the above relations to be well defined, I've determined
that $0<x<log_4(35)$ and $0<y<log_3(11)$. I also know that each of the equations(taken separately by itself) has exactly one solution because of their monotonicity (in each case one is strictly increasing and the other is strictly decreasing, so they can't have more than one intersection.)




By substituting $(1-log_{10}(5))$ with $log_{10}(2)$ in the first equation the unique solution for it is found to be $x=log_2(11-3^y).$


I've tried to proceed further but to no avail and come here looking for help; I think that there is no real solution to the system but haven't been able to prove it.






algebra-precalculus logarithms exponential-function systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Base of the logarithm $lg$?
    $endgroup$
    – Robert Z
    Feb 16 at 11:22










  • $begingroup$
    10 is the base of lg of course
    $endgroup$
    – Luca Pana
    Feb 16 at 11:22














3












3








3





$begingroup$



Solve the following equation system over the real numbers
$$begin{cases}
x(1-log_{10}(5))=log_{10}(11-3^y)\
log_{10}(35-4^x)=ylog_{10}(9) \
end{cases}
$$





For the functions in the above relations to be well defined, I've determined
that $0<x<log_4(35)$ and $0<y<log_3(11)$. I also know that each of the equations(taken separately by itself) has exactly one solution because of their monotonicity (in each case one is strictly increasing and the other is strictly decreasing, so they can't have more than one intersection.)




By substituting $(1-log_{10}(5))$ with $log_{10}(2)$ in the first equation the unique solution for it is found to be $x=log_2(11-3^y).$


I've tried to proceed further but to no avail and come here looking for help; I think that there is no real solution to the system but haven't been able to prove it.






algebra-precalculus logarithms exponential-function systems-of-equations







share|cite|improve this question











$endgroup$





Solve the following equation system over the real numbers
$$begin{cases}
x(1-log_{10}(5))=log_{10}(11-3^y)\
log_{10}(35-4^x)=ylog_{10}(9) \
end{cases}
$$





For the functions in the above relations to be well defined, I've determined
that $0<x<log_4(35)$ and $0<y<log_3(11)$. I also know that each of the equations(taken separately by itself) has exactly one solution because of their monotonicity (in each case one is strictly increasing and the other is strictly decreasing, so they can't have more than one intersection.)




By substituting $(1-log_{10}(5))$ with $log_{10}(2)$ in the first equation the unique solution for it is found to be $x=log_2(11-3^y).$


I've tried to proceed further but to no avail and come here looking for help; I think that there is no real solution to the system but haven't been able to prove it.






algebra-precalculus logarithms exponential-function systems-of-equations




algebra-precalculus logarithms exponential-function systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 16 at 11:25









Robert Z

98.6k1068139




98.6k1068139










asked Feb 16 at 11:19









Luca PanaLuca Pana

1197




1197












  • $begingroup$
    Base of the logarithm $lg$?
    $endgroup$
    – Robert Z
    Feb 16 at 11:22










  • $begingroup$
    10 is the base of lg of course
    $endgroup$
    – Luca Pana
    Feb 16 at 11:22


















  • $begingroup$
    Base of the logarithm $lg$?
    $endgroup$
    – Robert Z
    Feb 16 at 11:22










  • $begingroup$
    10 is the base of lg of course
    $endgroup$
    – Luca Pana
    Feb 16 at 11:22
















$begingroup$
Base of the logarithm $lg$?
$endgroup$
– Robert Z
Feb 16 at 11:22




$begingroup$
Base of the logarithm $lg$?
$endgroup$
– Robert Z
Feb 16 at 11:22












$begingroup$
10 is the base of lg of course
$endgroup$
– Luca Pana
Feb 16 at 11:22




$begingroup$
10 is the base of lg of course
$endgroup$
– Luca Pana
Feb 16 at 11:22










3 Answers
3






active

oldest

votes


















7












$begingroup$

The system can be written as
$$begin{cases}
log_{10}(2^x)=log_{10}(11-3^y)\
log_{10}(35-(2^{x})^2)=log_{10}((3^{y})^2)
end{cases}$$
Now if $u:=2^x<sqrt{35}$ and $v:=3^y<11$, we can throw away the logarithm and solve with respect to $u$ and $v$:
$$begin{cases}
u+v=11\
u^2+v^2=35
end{cases}$$
Can you take it from here?



P.S. Yes, as you already noted there are no real solutions otherwise
$$70=2(u^2+v^2)=(u+v)^2+(u-v)^2geq(u+v)^2=121.$$






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Hint: It is $$2^x=11-3^y$$ and $$35-2^{2x}=3^{2y}$$ Can you proceed?






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Hint



      We have that



      $$x(1-log(5))=log(11-3^y)iff 2^x=11-3^y$$ and



      $$log(35-4^x)=ylog(9) iff 35-4^x=9^y.$$






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

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        7












        $begingroup$

        The system can be written as
        $$begin{cases}
        log_{10}(2^x)=log_{10}(11-3^y)\
        log_{10}(35-(2^{x})^2)=log_{10}((3^{y})^2)
        end{cases}$$
        Now if $u:=2^x<sqrt{35}$ and $v:=3^y<11$, we can throw away the logarithm and solve with respect to $u$ and $v$:
        $$begin{cases}
        u+v=11\
        u^2+v^2=35
        end{cases}$$
        Can you take it from here?



        P.S. Yes, as you already noted there are no real solutions otherwise
        $$70=2(u^2+v^2)=(u+v)^2+(u-v)^2geq(u+v)^2=121.$$






        share|cite|improve this answer











        $endgroup$


















          7












          $begingroup$

          The system can be written as
          $$begin{cases}
          log_{10}(2^x)=log_{10}(11-3^y)\
          log_{10}(35-(2^{x})^2)=log_{10}((3^{y})^2)
          end{cases}$$
          Now if $u:=2^x<sqrt{35}$ and $v:=3^y<11$, we can throw away the logarithm and solve with respect to $u$ and $v$:
          $$begin{cases}
          u+v=11\
          u^2+v^2=35
          end{cases}$$
          Can you take it from here?



          P.S. Yes, as you already noted there are no real solutions otherwise
          $$70=2(u^2+v^2)=(u+v)^2+(u-v)^2geq(u+v)^2=121.$$






          share|cite|improve this answer











          $endgroup$
















            7












            7








            7





            $begingroup$

            The system can be written as
            $$begin{cases}
            log_{10}(2^x)=log_{10}(11-3^y)\
            log_{10}(35-(2^{x})^2)=log_{10}((3^{y})^2)
            end{cases}$$
            Now if $u:=2^x<sqrt{35}$ and $v:=3^y<11$, we can throw away the logarithm and solve with respect to $u$ and $v$:
            $$begin{cases}
            u+v=11\
            u^2+v^2=35
            end{cases}$$
            Can you take it from here?



            P.S. Yes, as you already noted there are no real solutions otherwise
            $$70=2(u^2+v^2)=(u+v)^2+(u-v)^2geq(u+v)^2=121.$$






            share|cite|improve this answer











            $endgroup$



            The system can be written as
            $$begin{cases}
            log_{10}(2^x)=log_{10}(11-3^y)\
            log_{10}(35-(2^{x})^2)=log_{10}((3^{y})^2)
            end{cases}$$
            Now if $u:=2^x<sqrt{35}$ and $v:=3^y<11$, we can throw away the logarithm and solve with respect to $u$ and $v$:
            $$begin{cases}
            u+v=11\
            u^2+v^2=35
            end{cases}$$
            Can you take it from here?



            P.S. Yes, as you already noted there are no real solutions otherwise
            $$70=2(u^2+v^2)=(u+v)^2+(u-v)^2geq(u+v)^2=121.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 16 at 12:26

























            answered Feb 16 at 11:31









            Robert ZRobert Z

            98.6k1068139




            98.6k1068139























                3












                $begingroup$

                Hint: It is $$2^x=11-3^y$$ and $$35-2^{2x}=3^{2y}$$ Can you proceed?






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Hint: It is $$2^x=11-3^y$$ and $$35-2^{2x}=3^{2y}$$ Can you proceed?






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Hint: It is $$2^x=11-3^y$$ and $$35-2^{2x}=3^{2y}$$ Can you proceed?






                    share|cite|improve this answer









                    $endgroup$



                    Hint: It is $$2^x=11-3^y$$ and $$35-2^{2x}=3^{2y}$$ Can you proceed?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 16 at 11:27









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    76.2k42866




                    76.2k42866























                        1












                        $begingroup$

                        Hint



                        We have that



                        $$x(1-log(5))=log(11-3^y)iff 2^x=11-3^y$$ and



                        $$log(35-4^x)=ylog(9) iff 35-4^x=9^y.$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint



                          We have that



                          $$x(1-log(5))=log(11-3^y)iff 2^x=11-3^y$$ and



                          $$log(35-4^x)=ylog(9) iff 35-4^x=9^y.$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint



                            We have that



                            $$x(1-log(5))=log(11-3^y)iff 2^x=11-3^y$$ and



                            $$log(35-4^x)=ylog(9) iff 35-4^x=9^y.$$






                            share|cite|improve this answer









                            $endgroup$



                            Hint



                            We have that



                            $$x(1-log(5))=log(11-3^y)iff 2^x=11-3^y$$ and



                            $$log(35-4^x)=ylog(9) iff 35-4^x=9^y.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 16 at 11:24









                            mflmfl

                            26.6k12142




                            26.6k12142






























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