How do I find the solutions of $|x-2|^{10x^2-1}=|x-2|^{3x}$?












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How do I find the solutions of the following equation: $$|x-2|^{10x^2-1}=|x-2|^{3x}?$$




It has 5 solutions, 4 positive and 1 negative. The graphs are these.



How do I compute the values of these roots manually?






algebra-precalculus logarithms







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  • $begingroup$
    math.stackexchange.com/questions/3157637/…
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    – lab bhattacharjee
    Mar 27 at 19:27
















1












$begingroup$



How do I find the solutions of the following equation: $$|x-2|^{10x^2-1}=|x-2|^{3x}?$$




It has 5 solutions, 4 positive and 1 negative. The graphs are these.



How do I compute the values of these roots manually?






algebra-precalculus logarithms







share|cite|improve this question









New contributor




Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/3157637/…
    $endgroup$
    – lab bhattacharjee
    Mar 27 at 19:27














1












1








1


2



$begingroup$



How do I find the solutions of the following equation: $$|x-2|^{10x^2-1}=|x-2|^{3x}?$$




It has 5 solutions, 4 positive and 1 negative. The graphs are these.



How do I compute the values of these roots manually?






algebra-precalculus logarithms







share|cite|improve this question









New contributor




Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





How do I find the solutions of the following equation: $$|x-2|^{10x^2-1}=|x-2|^{3x}?$$




It has 5 solutions, 4 positive and 1 negative. The graphs are these.



How do I compute the values of these roots manually?






algebra-precalculus logarithms




algebra-precalculus logarithms






share|cite|improve this question









New contributor




Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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edited Mar 28 at 19:00









Maria Mazur

49.5k1361124




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asked Mar 27 at 18:36









Namami ShankerNamami Shanker

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  • $begingroup$
    math.stackexchange.com/questions/3157637/…
    $endgroup$
    – lab bhattacharjee
    Mar 27 at 19:27


















  • $begingroup$
    math.stackexchange.com/questions/3157637/…
    $endgroup$
    – lab bhattacharjee
    Mar 27 at 19:27
















$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 27 at 19:27




$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
Mar 27 at 19:27










4 Answers
4






active

oldest

votes


















12












$begingroup$

We see that $x=2$ is one solution. Let $xne 2$.



Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$



So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.



Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= {1over 2}$ and $x=-{1over 5}$.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    So rearranging gives
    $$|x-2|^{10x^2-1}-|x-2|^{3x}=0$$
    $$|x-2|^{3x}(|x-2|^{10x^2-3x-1}-1)=0$$
    So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Hint



      Either $$x=2$$or$$|x-2|^{10x^2-3x-1}=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Yes thank you sir.
          $endgroup$
          – Namami Shanker
          Mar 27 at 18:45










        • $begingroup$
          This does not give all of the solutions.
          $endgroup$
          – Peter Foreman
          Mar 27 at 18:47










        • $begingroup$
          The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
          $endgroup$
          – Robert Israel
          Mar 27 at 18:54














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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

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        active

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        active

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        12












        $begingroup$

        We see that $x=2$ is one solution. Let $xne 2$.



        Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$



        So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.



        Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= {1over 2}$ and $x=-{1over 5}$.






        share|cite|improve this answer











        $endgroup$


















          12












          $begingroup$

          We see that $x=2$ is one solution. Let $xne 2$.



          Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$



          So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.



          Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= {1over 2}$ and $x=-{1over 5}$.






          share|cite|improve this answer











          $endgroup$
















            12












            12








            12





            $begingroup$

            We see that $x=2$ is one solution. Let $xne 2$.



            Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$



            So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.



            Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= {1over 2}$ and $x=-{1over 5}$.






            share|cite|improve this answer











            $endgroup$



            We see that $x=2$ is one solution. Let $xne 2$.



            Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$



            So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.



            Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= {1over 2}$ and $x=-{1over 5}$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 27 at 20:39









            Moo

            5,64031020




            5,64031020










            answered Mar 27 at 18:43









            Maria MazurMaria Mazur

            49.5k1361124




            49.5k1361124























                4












                $begingroup$

                So rearranging gives
                $$|x-2|^{10x^2-1}-|x-2|^{3x}=0$$
                $$|x-2|^{3x}(|x-2|^{10x^2-3x-1}-1)=0$$
                So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  So rearranging gives
                  $$|x-2|^{10x^2-1}-|x-2|^{3x}=0$$
                  $$|x-2|^{3x}(|x-2|^{10x^2-3x-1}-1)=0$$
                  So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    So rearranging gives
                    $$|x-2|^{10x^2-1}-|x-2|^{3x}=0$$
                    $$|x-2|^{3x}(|x-2|^{10x^2-3x-1}-1)=0$$
                    So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.






                    share|cite|improve this answer









                    $endgroup$



                    So rearranging gives
                    $$|x-2|^{10x^2-1}-|x-2|^{3x}=0$$
                    $$|x-2|^{3x}(|x-2|^{10x^2-3x-1}-1)=0$$
                    So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 27 at 18:45









                    Peter ForemanPeter Foreman

                    5,5391216




                    5,5391216























                        1












                        $begingroup$

                        Hint



                        Either $$x=2$$or$$|x-2|^{10x^2-3x-1}=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint



                          Either $$x=2$$or$$|x-2|^{10x^2-3x-1}=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint



                            Either $$x=2$$or$$|x-2|^{10x^2-3x-1}=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)






                            share|cite|improve this answer









                            $endgroup$



                            Hint



                            Either $$x=2$$or$$|x-2|^{10x^2-3x-1}=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 27 at 19:07









                            Mostafa AyazMostafa Ayaz

                            18.2k31040




                            18.2k31040























                                0












                                $begingroup$

                                We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Yes thank you sir.
                                  $endgroup$
                                  – Namami Shanker
                                  Mar 27 at 18:45










                                • $begingroup$
                                  This does not give all of the solutions.
                                  $endgroup$
                                  – Peter Foreman
                                  Mar 27 at 18:47










                                • $begingroup$
                                  The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                                  $endgroup$
                                  – Robert Israel
                                  Mar 27 at 18:54


















                                0












                                $begingroup$

                                We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Yes thank you sir.
                                  $endgroup$
                                  – Namami Shanker
                                  Mar 27 at 18:45










                                • $begingroup$
                                  This does not give all of the solutions.
                                  $endgroup$
                                  – Peter Foreman
                                  Mar 27 at 18:47










                                • $begingroup$
                                  The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                                  $endgroup$
                                  – Robert Israel
                                  Mar 27 at 18:54
















                                0












                                0








                                0





                                $begingroup$

                                We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.






                                share|cite|improve this answer











                                $endgroup$



                                We get easy that $x=2$ is one solution. Now let $xneq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Mar 28 at 2:08









                                Solomon Ucko

                                14719




                                14719










                                answered Mar 27 at 18:41









                                Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                                78.4k42867




                                78.4k42867












                                • $begingroup$
                                  Yes thank you sir.
                                  $endgroup$
                                  – Namami Shanker
                                  Mar 27 at 18:45










                                • $begingroup$
                                  This does not give all of the solutions.
                                  $endgroup$
                                  – Peter Foreman
                                  Mar 27 at 18:47










                                • $begingroup$
                                  The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                                  $endgroup$
                                  – Robert Israel
                                  Mar 27 at 18:54




















                                • $begingroup$
                                  Yes thank you sir.
                                  $endgroup$
                                  – Namami Shanker
                                  Mar 27 at 18:45










                                • $begingroup$
                                  This does not give all of the solutions.
                                  $endgroup$
                                  – Peter Foreman
                                  Mar 27 at 18:47










                                • $begingroup$
                                  The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                                  $endgroup$
                                  – Robert Israel
                                  Mar 27 at 18:54


















                                $begingroup$
                                Yes thank you sir.
                                $endgroup$
                                – Namami Shanker
                                Mar 27 at 18:45




                                $begingroup$
                                Yes thank you sir.
                                $endgroup$
                                – Namami Shanker
                                Mar 27 at 18:45












                                $begingroup$
                                This does not give all of the solutions.
                                $endgroup$
                                – Peter Foreman
                                Mar 27 at 18:47




                                $begingroup$
                                This does not give all of the solutions.
                                $endgroup$
                                – Peter Foreman
                                Mar 27 at 18:47












                                $begingroup$
                                The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                                $endgroup$
                                – Robert Israel
                                Mar 27 at 18:54






                                $begingroup$
                                The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                                $endgroup$
                                – Robert Israel
                                Mar 27 at 18:54












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                                Cáceres

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                                Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
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