Why is the solution to the equation $ln(x)+ln(3x+1) = 0$ not $x=frac{-1}{4}$?












-1












$begingroup$


Wouldn't you just move the $ln(x)$ to the other side, then raise both sides to the power of $e$? And then you have the same bases equal to each other, so you get $3x + 1 = x$? From where you'd get $x =frac{ -1}{4}$? Is there something wrong with this process?



This question was on a calculus exam so it does seem overly simple...any help is appreciated.






calculus algebra-precalculus







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  • 1




    $begingroup$
    FYI, the solution to $3x + 1 = x$ is $x = -1/2$.
    $endgroup$
    – eyeballfrog
    Dec 7 '18 at 23:55


















-1












$begingroup$


Wouldn't you just move the $ln(x)$ to the other side, then raise both sides to the power of $e$? And then you have the same bases equal to each other, so you get $3x + 1 = x$? From where you'd get $x =frac{ -1}{4}$? Is there something wrong with this process?



This question was on a calculus exam so it does seem overly simple...any help is appreciated.






calculus algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    FYI, the solution to $3x + 1 = x$ is $x = -1/2$.
    $endgroup$
    – eyeballfrog
    Dec 7 '18 at 23:55
















-1












-1








-1





$begingroup$


Wouldn't you just move the $ln(x)$ to the other side, then raise both sides to the power of $e$? And then you have the same bases equal to each other, so you get $3x + 1 = x$? From where you'd get $x =frac{ -1}{4}$? Is there something wrong with this process?



This question was on a calculus exam so it does seem overly simple...any help is appreciated.






calculus algebra-precalculus







share|cite|improve this question











$endgroup$




Wouldn't you just move the $ln(x)$ to the other side, then raise both sides to the power of $e$? And then you have the same bases equal to each other, so you get $3x + 1 = x$? From where you'd get $x =frac{ -1}{4}$? Is there something wrong with this process?



This question was on a calculus exam so it does seem overly simple...any help is appreciated.






calculus algebra-precalculus




calculus algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 7:54









gimusi

92.8k84494




92.8k84494










asked Dec 7 '18 at 23:48









James RonaldJames Ronald

1257




1257








  • 1




    $begingroup$
    FYI, the solution to $3x + 1 = x$ is $x = -1/2$.
    $endgroup$
    – eyeballfrog
    Dec 7 '18 at 23:55
















  • 1




    $begingroup$
    FYI, the solution to $3x + 1 = x$ is $x = -1/2$.
    $endgroup$
    – eyeballfrog
    Dec 7 '18 at 23:55










1




1




$begingroup$
FYI, the solution to $3x + 1 = x$ is $x = -1/2$.
$endgroup$
– eyeballfrog
Dec 7 '18 at 23:55






$begingroup$
FYI, the solution to $3x + 1 = x$ is $x = -1/2$.
$endgroup$
– eyeballfrog
Dec 7 '18 at 23:55












4 Answers
4






active

oldest

votes


















5












$begingroup$

We don't need to use esponential, indeed by $log A+ log B = log (AB)$ with the conditions $x>0 implies 3x+1>0$, we have that



$$ln x + ln(3x+1)=ln (3x^2+x)=0 iff 3x^2+x=1 iff 3x^2+x-1=0$$



then



$$x=frac{-1pm sqrt{1+12}}{6} implies x=frac{-1+ sqrt{13}}{6}$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Yes, there is and no, that's not the first thing you do. You should first impose the condition that both $x$ and $3x+1$ are elements of the domain of $ln$, and then do any algebra you desire.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      I think you made a mistake of signs when you moved the $ln x$ to the other side. It should be $ln(3x+1)=-ln(x)$.



      Now you exponentiate both sides, and get $3x+1=frac1x$.



      After you're done solving that equation, remember to check that those solutions still work with the original equation; logarithms do not like negative inputs. Or you can do what the other answer says and keep track of which values of $x$ are valid as you go, and check your solutions against that at the end.






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        If you do that you get $ln x = -ln (3x+1)$



        and when you use both sides as powers of $e$ to get



        $e^{ln x} = e^{-ln (3x+1)} = frac 1{e^{3x+1}}$ so that your first mistake. Because one side negative you will get a reciprical.



        $x =frac 1{3x+1}$ so



        $x(3x+1) = 1$



        $3x^2 + x -1=0$



        So $x = frac {-1 pm sqrt{13}}6$ but $x > 0$ as logarithms of non-positives are not defined.



        So $x = frac {-1 + sqrt{13}}6$



        Alternatively you could have combined to get



        $ln x + ln(x+3) = 0$



        $ln x(x+3) = 0$ so



        $x(x+3) = 1$ and etc.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          The solutions are wrong
          $endgroup$
          – AryanSonwatikar
          Dec 8 '18 at 0:53










        • $begingroup$
          Knew something was weird.....
          $endgroup$
          – fleablood
          Dec 8 '18 at 1:18










        • $begingroup$
          @fleablood Now its fine.
          $endgroup$
          – gimusi
          Dec 8 '18 at 7:10











        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        We don't need to use esponential, indeed by $log A+ log B = log (AB)$ with the conditions $x>0 implies 3x+1>0$, we have that



        $$ln x + ln(3x+1)=ln (3x^2+x)=0 iff 3x^2+x=1 iff 3x^2+x-1=0$$



        then



        $$x=frac{-1pm sqrt{1+12}}{6} implies x=frac{-1+ sqrt{13}}{6}$$






        share|cite|improve this answer











        $endgroup$


















          5












          $begingroup$

          We don't need to use esponential, indeed by $log A+ log B = log (AB)$ with the conditions $x>0 implies 3x+1>0$, we have that



          $$ln x + ln(3x+1)=ln (3x^2+x)=0 iff 3x^2+x=1 iff 3x^2+x-1=0$$



          then



          $$x=frac{-1pm sqrt{1+12}}{6} implies x=frac{-1+ sqrt{13}}{6}$$






          share|cite|improve this answer











          $endgroup$
















            5












            5








            5





            $begingroup$

            We don't need to use esponential, indeed by $log A+ log B = log (AB)$ with the conditions $x>0 implies 3x+1>0$, we have that



            $$ln x + ln(3x+1)=ln (3x^2+x)=0 iff 3x^2+x=1 iff 3x^2+x-1=0$$



            then



            $$x=frac{-1pm sqrt{1+12}}{6} implies x=frac{-1+ sqrt{13}}{6}$$






            share|cite|improve this answer











            $endgroup$



            We don't need to use esponential, indeed by $log A+ log B = log (AB)$ with the conditions $x>0 implies 3x+1>0$, we have that



            $$ln x + ln(3x+1)=ln (3x^2+x)=0 iff 3x^2+x=1 iff 3x^2+x-1=0$$



            then



            $$x=frac{-1pm sqrt{1+12}}{6} implies x=frac{-1+ sqrt{13}}{6}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 8 '18 at 0:06

























            answered Dec 8 '18 at 0:00









            gimusigimusi

            92.8k84494




            92.8k84494























                1












                $begingroup$

                Yes, there is and no, that's not the first thing you do. You should first impose the condition that both $x$ and $3x+1$ are elements of the domain of $ln$, and then do any algebra you desire.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Yes, there is and no, that's not the first thing you do. You should first impose the condition that both $x$ and $3x+1$ are elements of the domain of $ln$, and then do any algebra you desire.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Yes, there is and no, that's not the first thing you do. You should first impose the condition that both $x$ and $3x+1$ are elements of the domain of $ln$, and then do any algebra you desire.






                    share|cite|improve this answer









                    $endgroup$



                    Yes, there is and no, that's not the first thing you do. You should first impose the condition that both $x$ and $3x+1$ are elements of the domain of $ln$, and then do any algebra you desire.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 7 '18 at 23:50









                    Saucy O'PathSaucy O'Path

                    5,9641627




                    5,9641627























                        1












                        $begingroup$

                        I think you made a mistake of signs when you moved the $ln x$ to the other side. It should be $ln(3x+1)=-ln(x)$.



                        Now you exponentiate both sides, and get $3x+1=frac1x$.



                        After you're done solving that equation, remember to check that those solutions still work with the original equation; logarithms do not like negative inputs. Or you can do what the other answer says and keep track of which values of $x$ are valid as you go, and check your solutions against that at the end.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          I think you made a mistake of signs when you moved the $ln x$ to the other side. It should be $ln(3x+1)=-ln(x)$.



                          Now you exponentiate both sides, and get $3x+1=frac1x$.



                          After you're done solving that equation, remember to check that those solutions still work with the original equation; logarithms do not like negative inputs. Or you can do what the other answer says and keep track of which values of $x$ are valid as you go, and check your solutions against that at the end.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            I think you made a mistake of signs when you moved the $ln x$ to the other side. It should be $ln(3x+1)=-ln(x)$.



                            Now you exponentiate both sides, and get $3x+1=frac1x$.



                            After you're done solving that equation, remember to check that those solutions still work with the original equation; logarithms do not like negative inputs. Or you can do what the other answer says and keep track of which values of $x$ are valid as you go, and check your solutions against that at the end.






                            share|cite|improve this answer









                            $endgroup$



                            I think you made a mistake of signs when you moved the $ln x$ to the other side. It should be $ln(3x+1)=-ln(x)$.



                            Now you exponentiate both sides, and get $3x+1=frac1x$.



                            After you're done solving that equation, remember to check that those solutions still work with the original equation; logarithms do not like negative inputs. Or you can do what the other answer says and keep track of which values of $x$ are valid as you go, and check your solutions against that at the end.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 7 '18 at 23:54









                            ArthurArthur

                            114k7115197




                            114k7115197























                                1












                                $begingroup$

                                If you do that you get $ln x = -ln (3x+1)$



                                and when you use both sides as powers of $e$ to get



                                $e^{ln x} = e^{-ln (3x+1)} = frac 1{e^{3x+1}}$ so that your first mistake. Because one side negative you will get a reciprical.



                                $x =frac 1{3x+1}$ so



                                $x(3x+1) = 1$



                                $3x^2 + x -1=0$



                                So $x = frac {-1 pm sqrt{13}}6$ but $x > 0$ as logarithms of non-positives are not defined.



                                So $x = frac {-1 + sqrt{13}}6$



                                Alternatively you could have combined to get



                                $ln x + ln(x+3) = 0$



                                $ln x(x+3) = 0$ so



                                $x(x+3) = 1$ and etc.






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  The solutions are wrong
                                  $endgroup$
                                  – AryanSonwatikar
                                  Dec 8 '18 at 0:53










                                • $begingroup$
                                  Knew something was weird.....
                                  $endgroup$
                                  – fleablood
                                  Dec 8 '18 at 1:18










                                • $begingroup$
                                  @fleablood Now its fine.
                                  $endgroup$
                                  – gimusi
                                  Dec 8 '18 at 7:10
















                                1












                                $begingroup$

                                If you do that you get $ln x = -ln (3x+1)$



                                and when you use both sides as powers of $e$ to get



                                $e^{ln x} = e^{-ln (3x+1)} = frac 1{e^{3x+1}}$ so that your first mistake. Because one side negative you will get a reciprical.



                                $x =frac 1{3x+1}$ so



                                $x(3x+1) = 1$



                                $3x^2 + x -1=0$



                                So $x = frac {-1 pm sqrt{13}}6$ but $x > 0$ as logarithms of non-positives are not defined.



                                So $x = frac {-1 + sqrt{13}}6$



                                Alternatively you could have combined to get



                                $ln x + ln(x+3) = 0$



                                $ln x(x+3) = 0$ so



                                $x(x+3) = 1$ and etc.






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  The solutions are wrong
                                  $endgroup$
                                  – AryanSonwatikar
                                  Dec 8 '18 at 0:53










                                • $begingroup$
                                  Knew something was weird.....
                                  $endgroup$
                                  – fleablood
                                  Dec 8 '18 at 1:18










                                • $begingroup$
                                  @fleablood Now its fine.
                                  $endgroup$
                                  – gimusi
                                  Dec 8 '18 at 7:10














                                1












                                1








                                1





                                $begingroup$

                                If you do that you get $ln x = -ln (3x+1)$



                                and when you use both sides as powers of $e$ to get



                                $e^{ln x} = e^{-ln (3x+1)} = frac 1{e^{3x+1}}$ so that your first mistake. Because one side negative you will get a reciprical.



                                $x =frac 1{3x+1}$ so



                                $x(3x+1) = 1$



                                $3x^2 + x -1=0$



                                So $x = frac {-1 pm sqrt{13}}6$ but $x > 0$ as logarithms of non-positives are not defined.



                                So $x = frac {-1 + sqrt{13}}6$



                                Alternatively you could have combined to get



                                $ln x + ln(x+3) = 0$



                                $ln x(x+3) = 0$ so



                                $x(x+3) = 1$ and etc.






                                share|cite|improve this answer











                                $endgroup$



                                If you do that you get $ln x = -ln (3x+1)$



                                and when you use both sides as powers of $e$ to get



                                $e^{ln x} = e^{-ln (3x+1)} = frac 1{e^{3x+1}}$ so that your first mistake. Because one side negative you will get a reciprical.



                                $x =frac 1{3x+1}$ so



                                $x(3x+1) = 1$



                                $3x^2 + x -1=0$



                                So $x = frac {-1 pm sqrt{13}}6$ but $x > 0$ as logarithms of non-positives are not defined.



                                So $x = frac {-1 + sqrt{13}}6$



                                Alternatively you could have combined to get



                                $ln x + ln(x+3) = 0$



                                $ln x(x+3) = 0$ so



                                $x(x+3) = 1$ and etc.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Dec 8 '18 at 7:09









                                gimusi

                                92.8k84494




                                92.8k84494










                                answered Dec 8 '18 at 0:14









                                fleabloodfleablood

                                70.3k22685




                                70.3k22685












                                • $begingroup$
                                  The solutions are wrong
                                  $endgroup$
                                  – AryanSonwatikar
                                  Dec 8 '18 at 0:53










                                • $begingroup$
                                  Knew something was weird.....
                                  $endgroup$
                                  – fleablood
                                  Dec 8 '18 at 1:18










                                • $begingroup$
                                  @fleablood Now its fine.
                                  $endgroup$
                                  – gimusi
                                  Dec 8 '18 at 7:10


















                                • $begingroup$
                                  The solutions are wrong
                                  $endgroup$
                                  – AryanSonwatikar
                                  Dec 8 '18 at 0:53










                                • $begingroup$
                                  Knew something was weird.....
                                  $endgroup$
                                  – fleablood
                                  Dec 8 '18 at 1:18










                                • $begingroup$
                                  @fleablood Now its fine.
                                  $endgroup$
                                  – gimusi
                                  Dec 8 '18 at 7:10
















                                $begingroup$
                                The solutions are wrong
                                $endgroup$
                                – AryanSonwatikar
                                Dec 8 '18 at 0:53




                                $begingroup$
                                The solutions are wrong
                                $endgroup$
                                – AryanSonwatikar
                                Dec 8 '18 at 0:53












                                $begingroup$
                                Knew something was weird.....
                                $endgroup$
                                – fleablood
                                Dec 8 '18 at 1:18




                                $begingroup$
                                Knew something was weird.....
                                $endgroup$
                                – fleablood
                                Dec 8 '18 at 1:18












                                $begingroup$
                                @fleablood Now its fine.
                                $endgroup$
                                – gimusi
                                Dec 8 '18 at 7:10




                                $begingroup$
                                @fleablood Now its fine.
                                $endgroup$
                                – gimusi
                                Dec 8 '18 at 7:10


















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