Solving $log_{2x}2 = log_{3x}3$, I find that $x$ can be any number. Did I do something wrong?












0












$begingroup$


Given the question




Find all possible values of $x$ when $log_{2x}2 = log_{3x}3$




This is how I did it..



$log_{2x}2 = log_{3x}3$



$1/ log_2 2x = 1/ log_3 3x$



Therefore $log_2 2x = log_3 3x$



$log_2 2 + log_2 x = log_3 3 + log_3 x$



$log_2 x = log_3 x$



$frac{log_2 x}{log_3 x} = 1$



$log_x x = 1$????



So x can be any number?



Can someone explain this a pre-calculus level? Thanks.






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You have asked many questions here - so many that you should know by now to use mathjax. Until you do I can't figure out what the question is. math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 1:47










  • $begingroup$
    @EthanBolker Hello, do you know how to make make two variables as your base of log in MathJax? I couldn't find it on that site
    $endgroup$
    – Ke Ke
    Dec 17 '18 at 1:49










  • $begingroup$
    use log_{2x} produces $log_{2x}$
    $endgroup$
    – Siong Thye Goh
    Dec 17 '18 at 1:50










  • $begingroup$
    OP: Can you confirm that my MathJax edit correctly captures your intention? If so, then your error is in transitioning from $frac{log_2 x}{log_3 x} = 1$ to $log_x x = 1$. I can't tell what you think you're doing there.
    $endgroup$
    – Brian Tung
    Dec 17 '18 at 1:52












  • $begingroup$
    Hint: mathwords.com/c/change_of_base_formula.htm
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 2:05
















0












$begingroup$


Given the question




Find all possible values of $x$ when $log_{2x}2 = log_{3x}3$




This is how I did it..



$log_{2x}2 = log_{3x}3$



$1/ log_2 2x = 1/ log_3 3x$



Therefore $log_2 2x = log_3 3x$



$log_2 2 + log_2 x = log_3 3 + log_3 x$



$log_2 x = log_3 x$



$frac{log_2 x}{log_3 x} = 1$



$log_x x = 1$????



So x can be any number?



Can someone explain this a pre-calculus level? Thanks.






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You have asked many questions here - so many that you should know by now to use mathjax. Until you do I can't figure out what the question is. math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 1:47










  • $begingroup$
    @EthanBolker Hello, do you know how to make make two variables as your base of log in MathJax? I couldn't find it on that site
    $endgroup$
    – Ke Ke
    Dec 17 '18 at 1:49










  • $begingroup$
    use log_{2x} produces $log_{2x}$
    $endgroup$
    – Siong Thye Goh
    Dec 17 '18 at 1:50










  • $begingroup$
    OP: Can you confirm that my MathJax edit correctly captures your intention? If so, then your error is in transitioning from $frac{log_2 x}{log_3 x} = 1$ to $log_x x = 1$. I can't tell what you think you're doing there.
    $endgroup$
    – Brian Tung
    Dec 17 '18 at 1:52












  • $begingroup$
    Hint: mathwords.com/c/change_of_base_formula.htm
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 2:05














0












0








0





$begingroup$


Given the question




Find all possible values of $x$ when $log_{2x}2 = log_{3x}3$




This is how I did it..



$log_{2x}2 = log_{3x}3$



$1/ log_2 2x = 1/ log_3 3x$



Therefore $log_2 2x = log_3 3x$



$log_2 2 + log_2 x = log_3 3 + log_3 x$



$log_2 x = log_3 x$



$frac{log_2 x}{log_3 x} = 1$



$log_x x = 1$????



So x can be any number?



Can someone explain this a pre-calculus level? Thanks.






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$




Given the question




Find all possible values of $x$ when $log_{2x}2 = log_{3x}3$




This is how I did it..



$log_{2x}2 = log_{3x}3$



$1/ log_2 2x = 1/ log_3 3x$



Therefore $log_2 2x = log_3 3x$



$log_2 2 + log_2 x = log_3 3 + log_3 x$



$log_2 x = log_3 x$



$frac{log_2 x}{log_3 x} = 1$



$log_x x = 1$????



So x can be any number?



Can someone explain this a pre-calculus level? Thanks.






algebra-precalculus logarithms




algebra-precalculus logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 1:56









Blue

49.1k870156




49.1k870156










asked Dec 17 '18 at 1:43









Ke KeKe Ke

1027




1027








  • 2




    $begingroup$
    You have asked many questions here - so many that you should know by now to use mathjax. Until you do I can't figure out what the question is. math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 1:47










  • $begingroup$
    @EthanBolker Hello, do you know how to make make two variables as your base of log in MathJax? I couldn't find it on that site
    $endgroup$
    – Ke Ke
    Dec 17 '18 at 1:49










  • $begingroup$
    use log_{2x} produces $log_{2x}$
    $endgroup$
    – Siong Thye Goh
    Dec 17 '18 at 1:50










  • $begingroup$
    OP: Can you confirm that my MathJax edit correctly captures your intention? If so, then your error is in transitioning from $frac{log_2 x}{log_3 x} = 1$ to $log_x x = 1$. I can't tell what you think you're doing there.
    $endgroup$
    – Brian Tung
    Dec 17 '18 at 1:52












  • $begingroup$
    Hint: mathwords.com/c/change_of_base_formula.htm
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 2:05














  • 2




    $begingroup$
    You have asked many questions here - so many that you should know by now to use mathjax. Until you do I can't figure out what the question is. math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 1:47










  • $begingroup$
    @EthanBolker Hello, do you know how to make make two variables as your base of log in MathJax? I couldn't find it on that site
    $endgroup$
    – Ke Ke
    Dec 17 '18 at 1:49










  • $begingroup$
    use log_{2x} produces $log_{2x}$
    $endgroup$
    – Siong Thye Goh
    Dec 17 '18 at 1:50










  • $begingroup$
    OP: Can you confirm that my MathJax edit correctly captures your intention? If so, then your error is in transitioning from $frac{log_2 x}{log_3 x} = 1$ to $log_x x = 1$. I can't tell what you think you're doing there.
    $endgroup$
    – Brian Tung
    Dec 17 '18 at 1:52












  • $begingroup$
    Hint: mathwords.com/c/change_of_base_formula.htm
    $endgroup$
    – Ethan Bolker
    Dec 17 '18 at 2:05








2




2




$begingroup$
You have asked many questions here - so many that you should know by now to use mathjax. Until you do I can't figure out what the question is. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 17 '18 at 1:47




$begingroup$
You have asked many questions here - so many that you should know by now to use mathjax. Until you do I can't figure out what the question is. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Ethan Bolker
Dec 17 '18 at 1:47












$begingroup$
@EthanBolker Hello, do you know how to make make two variables as your base of log in MathJax? I couldn't find it on that site
$endgroup$
– Ke Ke
Dec 17 '18 at 1:49




$begingroup$
@EthanBolker Hello, do you know how to make make two variables as your base of log in MathJax? I couldn't find it on that site
$endgroup$
– Ke Ke
Dec 17 '18 at 1:49












$begingroup$
use log_{2x} produces $log_{2x}$
$endgroup$
– Siong Thye Goh
Dec 17 '18 at 1:50




$begingroup$
use log_{2x} produces $log_{2x}$
$endgroup$
– Siong Thye Goh
Dec 17 '18 at 1:50












$begingroup$
OP: Can you confirm that my MathJax edit correctly captures your intention? If so, then your error is in transitioning from $frac{log_2 x}{log_3 x} = 1$ to $log_x x = 1$. I can't tell what you think you're doing there.
$endgroup$
– Brian Tung
Dec 17 '18 at 1:52






$begingroup$
OP: Can you confirm that my MathJax edit correctly captures your intention? If so, then your error is in transitioning from $frac{log_2 x}{log_3 x} = 1$ to $log_x x = 1$. I can't tell what you think you're doing there.
$endgroup$
– Brian Tung
Dec 17 '18 at 1:52














$begingroup$
Hint: mathwords.com/c/change_of_base_formula.htm
$endgroup$
– Ethan Bolker
Dec 17 '18 at 2:05




$begingroup$
Hint: mathwords.com/c/change_of_base_formula.htm
$endgroup$
– Ethan Bolker
Dec 17 '18 at 2:05










2 Answers
2






active

oldest

votes


















4












$begingroup$

I do not know how do you obtain $log_x(x)=1$.



From $log_2 (x) = log_3(x)$.



We have $$frac{ln x}{ln 2} = frac{ln x}{ln 3}$$



$$ln x left(frac1{ln 2} - frac1{ln 3} right)=0$$



Hence $ln x = 0$, hence $x=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hello, because any number like log_2(2) = 1 or log_1000(1000) = 1. So as long as the argument and base matches isn't it always 1?
    $endgroup$
    – Ke Ke
    Dec 17 '18 at 1:52










  • $begingroup$
    but how does that has to do with the question? what happens to $log_2$ and $log_3$?
    $endgroup$
    – Siong Thye Goh
    Dec 17 '18 at 1:54



















1












$begingroup$

How exactly did you perform the following step? It’s clearly not correct:



$$color{red}{frac{log_2 x}{log_3 x} = 1 iff log_x x = 1}$$



Instead, you must simplify as follows:



$$frac{log_2 x}{log_3 x} = 1 iff frac{frac{ln x}{ln 2}}{frac{ln x}{ln 3}} = 1 iff frac{ln x}{ln 2} = frac{ln x}{ln 3} iff x = 1$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    I do not know how do you obtain $log_x(x)=1$.



    From $log_2 (x) = log_3(x)$.



    We have $$frac{ln x}{ln 2} = frac{ln x}{ln 3}$$



    $$ln x left(frac1{ln 2} - frac1{ln 3} right)=0$$



    Hence $ln x = 0$, hence $x=1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hello, because any number like log_2(2) = 1 or log_1000(1000) = 1. So as long as the argument and base matches isn't it always 1?
      $endgroup$
      – Ke Ke
      Dec 17 '18 at 1:52










    • $begingroup$
      but how does that has to do with the question? what happens to $log_2$ and $log_3$?
      $endgroup$
      – Siong Thye Goh
      Dec 17 '18 at 1:54
















    4












    $begingroup$

    I do not know how do you obtain $log_x(x)=1$.



    From $log_2 (x) = log_3(x)$.



    We have $$frac{ln x}{ln 2} = frac{ln x}{ln 3}$$



    $$ln x left(frac1{ln 2} - frac1{ln 3} right)=0$$



    Hence $ln x = 0$, hence $x=1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Hello, because any number like log_2(2) = 1 or log_1000(1000) = 1. So as long as the argument and base matches isn't it always 1?
      $endgroup$
      – Ke Ke
      Dec 17 '18 at 1:52










    • $begingroup$
      but how does that has to do with the question? what happens to $log_2$ and $log_3$?
      $endgroup$
      – Siong Thye Goh
      Dec 17 '18 at 1:54














    4












    4








    4





    $begingroup$

    I do not know how do you obtain $log_x(x)=1$.



    From $log_2 (x) = log_3(x)$.



    We have $$frac{ln x}{ln 2} = frac{ln x}{ln 3}$$



    $$ln x left(frac1{ln 2} - frac1{ln 3} right)=0$$



    Hence $ln x = 0$, hence $x=1$.






    share|cite|improve this answer









    $endgroup$



    I do not know how do you obtain $log_x(x)=1$.



    From $log_2 (x) = log_3(x)$.



    We have $$frac{ln x}{ln 2} = frac{ln x}{ln 3}$$



    $$ln x left(frac1{ln 2} - frac1{ln 3} right)=0$$



    Hence $ln x = 0$, hence $x=1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 17 '18 at 1:49









    Siong Thye GohSiong Thye Goh

    103k1468119




    103k1468119












    • $begingroup$
      Hello, because any number like log_2(2) = 1 or log_1000(1000) = 1. So as long as the argument and base matches isn't it always 1?
      $endgroup$
      – Ke Ke
      Dec 17 '18 at 1:52










    • $begingroup$
      but how does that has to do with the question? what happens to $log_2$ and $log_3$?
      $endgroup$
      – Siong Thye Goh
      Dec 17 '18 at 1:54


















    • $begingroup$
      Hello, because any number like log_2(2) = 1 or log_1000(1000) = 1. So as long as the argument and base matches isn't it always 1?
      $endgroup$
      – Ke Ke
      Dec 17 '18 at 1:52










    • $begingroup$
      but how does that has to do with the question? what happens to $log_2$ and $log_3$?
      $endgroup$
      – Siong Thye Goh
      Dec 17 '18 at 1:54
















    $begingroup$
    Hello, because any number like log_2(2) = 1 or log_1000(1000) = 1. So as long as the argument and base matches isn't it always 1?
    $endgroup$
    – Ke Ke
    Dec 17 '18 at 1:52




    $begingroup$
    Hello, because any number like log_2(2) = 1 or log_1000(1000) = 1. So as long as the argument and base matches isn't it always 1?
    $endgroup$
    – Ke Ke
    Dec 17 '18 at 1:52












    $begingroup$
    but how does that has to do with the question? what happens to $log_2$ and $log_3$?
    $endgroup$
    – Siong Thye Goh
    Dec 17 '18 at 1:54




    $begingroup$
    but how does that has to do with the question? what happens to $log_2$ and $log_3$?
    $endgroup$
    – Siong Thye Goh
    Dec 17 '18 at 1:54











    1












    $begingroup$

    How exactly did you perform the following step? It’s clearly not correct:



    $$color{red}{frac{log_2 x}{log_3 x} = 1 iff log_x x = 1}$$



    Instead, you must simplify as follows:



    $$frac{log_2 x}{log_3 x} = 1 iff frac{frac{ln x}{ln 2}}{frac{ln x}{ln 3}} = 1 iff frac{ln x}{ln 2} = frac{ln x}{ln 3} iff x = 1$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      How exactly did you perform the following step? It’s clearly not correct:



      $$color{red}{frac{log_2 x}{log_3 x} = 1 iff log_x x = 1}$$



      Instead, you must simplify as follows:



      $$frac{log_2 x}{log_3 x} = 1 iff frac{frac{ln x}{ln 2}}{frac{ln x}{ln 3}} = 1 iff frac{ln x}{ln 2} = frac{ln x}{ln 3} iff x = 1$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        How exactly did you perform the following step? It’s clearly not correct:



        $$color{red}{frac{log_2 x}{log_3 x} = 1 iff log_x x = 1}$$



        Instead, you must simplify as follows:



        $$frac{log_2 x}{log_3 x} = 1 iff frac{frac{ln x}{ln 2}}{frac{ln x}{ln 3}} = 1 iff frac{ln x}{ln 2} = frac{ln x}{ln 3} iff x = 1$$






        share|cite|improve this answer









        $endgroup$



        How exactly did you perform the following step? It’s clearly not correct:



        $$color{red}{frac{log_2 x}{log_3 x} = 1 iff log_x x = 1}$$



        Instead, you must simplify as follows:



        $$frac{log_2 x}{log_3 x} = 1 iff frac{frac{ln x}{ln 2}}{frac{ln x}{ln 3}} = 1 iff frac{ln x}{ln 2} = frac{ln x}{ln 3} iff x = 1$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 4:26









        KM101KM101

        6,0901525




        6,0901525






























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