Simplify $dfraclog_3(12)log_36(3)-dfraclog_3(4)log_108(3)$ [closed]

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How simplify the logarithm?



$dfraclog_3(12)log_36(3)-dfraclog_3(4)log_108(3)$




logarithms fractions




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closed as off-topic by Leucippus, José Carlos Santos, heropup, Shailesh, Mathmo123 Aug 1 at 9:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, José Carlos Santos, heropup, Shailesh, Mathmo123
If this question can be reworded to fit the rules in the help center, please edit the question.












  • The step are: $$dfraclog_3(3cdot2^2)log_2^2cdot3^2(3)-dfraclog_3(2^2)log_2^2cdot3^3(3)$$ $$dfraclog_3(3)+2log_3(2)log_2^2cdot3^2(3)-dfrac2log_3(2)log_2^2cdot3^3(3)$$
    – leotv
    Aug 1 at 5:52















up vote
-1
down vote

favorite












How simplify the logarithm?



$dfraclog_3(12)log_36(3)-dfraclog_3(4)log_108(3)$




logarithms fractions




share|cite|improve this question













closed as off-topic by Leucippus, José Carlos Santos, heropup, Shailesh, Mathmo123 Aug 1 at 9:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, José Carlos Santos, heropup, Shailesh, Mathmo123
If this question can be reworded to fit the rules in the help center, please edit the question.












  • The step are: $$dfraclog_3(3cdot2^2)log_2^2cdot3^2(3)-dfraclog_3(2^2)log_2^2cdot3^3(3)$$ $$dfraclog_3(3)+2log_3(2)log_2^2cdot3^2(3)-dfrac2log_3(2)log_2^2cdot3^3(3)$$
    – leotv
    Aug 1 at 5:52













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











How simplify the logarithm?



$dfraclog_3(12)log_36(3)-dfraclog_3(4)log_108(3)$




logarithms fractions




share|cite|improve this question













How simplify the logarithm?



$dfraclog_3(12)log_36(3)-dfraclog_3(4)log_108(3)$





logarithms fractions





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 1 at 5:13
























asked Aug 1 at 5:10









leotv

11




11




closed as off-topic by Leucippus, José Carlos Santos, heropup, Shailesh, Mathmo123 Aug 1 at 9:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, José Carlos Santos, heropup, Shailesh, Mathmo123
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Leucippus, José Carlos Santos, heropup, Shailesh, Mathmo123 Aug 1 at 9:56


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, José Carlos Santos, heropup, Shailesh, Mathmo123
If this question can be reworded to fit the rules in the help center, please edit the question.











  • The step are: $$dfraclog_3(3cdot2^2)log_2^2cdot3^2(3)-dfraclog_3(2^2)log_2^2cdot3^3(3)$$ $$dfraclog_3(3)+2log_3(2)log_2^2cdot3^2(3)-dfrac2log_3(2)log_2^2cdot3^3(3)$$
    – leotv
    Aug 1 at 5:52

















  • The step are: $$dfraclog_3(3cdot2^2)log_2^2cdot3^2(3)-dfraclog_3(2^2)log_2^2cdot3^3(3)$$ $$dfraclog_3(3)+2log_3(2)log_2^2cdot3^2(3)-dfrac2log_3(2)log_2^2cdot3^3(3)$$
    – leotv
    Aug 1 at 5:52
















The step are: $$dfraclog_3(3cdot2^2)log_2^2cdot3^2(3)-dfraclog_3(2^2)log_2^2cdot3^3(3)$$ $$dfraclog_3(3)+2log_3(2)log_2^2cdot3^2(3)-dfrac2log_3(2)log_2^2cdot3^3(3)$$
– leotv
Aug 1 at 5:52





The step are: $$dfraclog_3(3cdot2^2)log_2^2cdot3^2(3)-dfraclog_3(2^2)log_2^2cdot3^3(3)$$ $$dfraclog_3(3)+2log_3(2)log_2^2cdot3^2(3)-dfrac2log_3(2)log_2^2cdot3^3(3)$$
– leotv
Aug 1 at 5:52











2 Answers
2






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up vote
0
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Let $log_32=a$.



Thus, it's
$$(1+2a)(2+2a)-2a(3+2a)=2.$$






share|cite|improve this answer























  • Why are you using $log_3(2)$?
    – leotv
    Aug 1 at 5:24










  • Because all our logarithms we can get by $log_32$.
    – Michael Rozenberg
    Aug 1 at 5:25

















up vote
0
down vote













From the perspective of a beginning algebra student, the easiest approach might be to convert everything to $log x$, that is $log_10x$ using the change of base relationship



$$ log_bx=fraclog xlog b $$



Then we have



begineqnarray
dfraclog_3(12)log_36(3)-dfraclog_3(4)log_108(3)&=&
fracfraclog12log3fraclog3log36-fracfraclog4log3fraclog3log108\
&=&fraclog12log36-log4log108(log3)^2\
&=&frac(log3+2log2)(2log3+2log2)-2log2(3log3+2log2)(log3)^2\
&=&frac2(log3)^2(log3)^2=2
endeqnarray






share|cite|improve this answer




























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote













    Let $log_32=a$.



    Thus, it's
    $$(1+2a)(2+2a)-2a(3+2a)=2.$$






    share|cite|improve this answer























    • Why are you using $log_3(2)$?
      – leotv
      Aug 1 at 5:24










    • Because all our logarithms we can get by $log_32$.
      – Michael Rozenberg
      Aug 1 at 5:25














    up vote
    0
    down vote













    Let $log_32=a$.



    Thus, it's
    $$(1+2a)(2+2a)-2a(3+2a)=2.$$






    share|cite|improve this answer























    • Why are you using $log_3(2)$?
      – leotv
      Aug 1 at 5:24










    • Because all our logarithms we can get by $log_32$.
      – Michael Rozenberg
      Aug 1 at 5:25












    up vote
    0
    down vote










    up vote
    0
    down vote









    Let $log_32=a$.



    Thus, it's
    $$(1+2a)(2+2a)-2a(3+2a)=2.$$






    share|cite|improve this answer















    Let $log_32=a$.



    Thus, it's
    $$(1+2a)(2+2a)-2a(3+2a)=2.$$







    share|cite|improve this answer















    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 1 at 5:20


























    answered Aug 1 at 5:13









    Michael Rozenberg

    87.4k1577179




    87.4k1577179











    • Why are you using $log_3(2)$?
      – leotv
      Aug 1 at 5:24










    • Because all our logarithms we can get by $log_32$.
      – Michael Rozenberg
      Aug 1 at 5:25
















    • Why are you using $log_3(2)$?
      – leotv
      Aug 1 at 5:24










    • Because all our logarithms we can get by $log_32$.
      – Michael Rozenberg
      Aug 1 at 5:25















    Why are you using $log_3(2)$?
    – leotv
    Aug 1 at 5:24




    Why are you using $log_3(2)$?
    – leotv
    Aug 1 at 5:24












    Because all our logarithms we can get by $log_32$.
    – Michael Rozenberg
    Aug 1 at 5:25




    Because all our logarithms we can get by $log_32$.
    – Michael Rozenberg
    Aug 1 at 5:25










    up vote
    0
    down vote













    From the perspective of a beginning algebra student, the easiest approach might be to convert everything to $log x$, that is $log_10x$ using the change of base relationship



    $$ log_bx=fraclog xlog b $$



    Then we have



    begineqnarray
    dfraclog_3(12)log_36(3)-dfraclog_3(4)log_108(3)&=&
    fracfraclog12log3fraclog3log36-fracfraclog4log3fraclog3log108\
    &=&fraclog12log36-log4log108(log3)^2\
    &=&frac(log3+2log2)(2log3+2log2)-2log2(3log3+2log2)(log3)^2\
    &=&frac2(log3)^2(log3)^2=2
    endeqnarray






    share|cite|improve this answer

























      up vote
      0
      down vote













      From the perspective of a beginning algebra student, the easiest approach might be to convert everything to $log x$, that is $log_10x$ using the change of base relationship



      $$ log_bx=fraclog xlog b $$



      Then we have



      begineqnarray
      dfraclog_3(12)log_36(3)-dfraclog_3(4)log_108(3)&=&
      fracfraclog12log3fraclog3log36-fracfraclog4log3fraclog3log108\
      &=&fraclog12log36-log4log108(log3)^2\
      &=&frac(log3+2log2)(2log3+2log2)-2log2(3log3+2log2)(log3)^2\
      &=&frac2(log3)^2(log3)^2=2
      endeqnarray






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        From the perspective of a beginning algebra student, the easiest approach might be to convert everything to $log x$, that is $log_10x$ using the change of base relationship



        $$ log_bx=fraclog xlog b $$



        Then we have



        begineqnarray
        dfraclog_3(12)log_36(3)-dfraclog_3(4)log_108(3)&=&
        fracfraclog12log3fraclog3log36-fracfraclog4log3fraclog3log108\
        &=&fraclog12log36-log4log108(log3)^2\
        &=&frac(log3+2log2)(2log3+2log2)-2log2(3log3+2log2)(log3)^2\
        &=&frac2(log3)^2(log3)^2=2
        endeqnarray






        share|cite|improve this answer













        From the perspective of a beginning algebra student, the easiest approach might be to convert everything to $log x$, that is $log_10x$ using the change of base relationship



        $$ log_bx=fraclog xlog b $$



        Then we have



        begineqnarray
        dfraclog_3(12)log_36(3)-dfraclog_3(4)log_108(3)&=&
        fracfraclog12log3fraclog3log36-fracfraclog4log3fraclog3log108\
        &=&fraclog12log36-log4log108(log3)^2\
        &=&frac(log3+2log2)(2log3+2log2)-2log2(3log3+2log2)(log3)^2\
        &=&frac2(log3)^2(log3)^2=2
        endeqnarray







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 1 at 7:14









        John Wayland Bales

        12.8k21135




        12.8k21135












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