How do I solve this logarithm problem? [closed]

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I'm trying to solve this problem:




If $log_27(a)=b$, find $log_sqrt[6]asqrt3$




However, I'm unable to see any connection in those given information. How can I solve this logarithm?




logarithms




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closed as off-topic by Henrik, John Ma, Xander Henderson, amWhy, user223391 Jul 31 at 1:08


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." – Henrik, John Ma, Xander Henderson, amWhy, Community
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    I'm unable to see any connection in those given information What do you know about logarithms change of base?
    – dxiv
    Jul 29 at 6:58










  • @dxiv you mean this notation: $log_ab=frac1log_ba$?
    – Steve
    Jul 29 at 7:03










  • @Steve: That's not notation, that's a formula - that seems relevant in this case - but I would just use $log_a b=fraclog blog a$ (from which your formula follows).
    – Henrik
    Jul 29 at 7:07






  • 2




    @Steve And more generally $log _ab=frac log _cblog _ca.,$
    – dxiv
    Jul 29 at 7:09






  • 1




    Why don't you just solve for $a$ and substitute?
    – Miksu
    Jul 29 at 7:58














up vote
1
down vote

favorite












I'm trying to solve this problem:




If $log_27(a)=b$, find $log_sqrt[6]asqrt3$




However, I'm unable to see any connection in those given information. How can I solve this logarithm?




logarithms




share|cite|improve this question













closed as off-topic by Henrik, John Ma, Xander Henderson, amWhy, user223391 Jul 31 at 1:08


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." – Henrik, John Ma, Xander Henderson, amWhy, Community
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    I'm unable to see any connection in those given information What do you know about logarithms change of base?
    – dxiv
    Jul 29 at 6:58










  • @dxiv you mean this notation: $log_ab=frac1log_ba$?
    – Steve
    Jul 29 at 7:03










  • @Steve: That's not notation, that's a formula - that seems relevant in this case - but I would just use $log_a b=fraclog blog a$ (from which your formula follows).
    – Henrik
    Jul 29 at 7:07






  • 2




    @Steve And more generally $log _ab=frac log _cblog _ca.,$
    – dxiv
    Jul 29 at 7:09






  • 1




    Why don't you just solve for $a$ and substitute?
    – Miksu
    Jul 29 at 7:58












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm trying to solve this problem:




If $log_27(a)=b$, find $log_sqrt[6]asqrt3$




However, I'm unable to see any connection in those given information. How can I solve this logarithm?




logarithms




share|cite|improve this question













I'm trying to solve this problem:




If $log_27(a)=b$, find $log_sqrt[6]asqrt3$




However, I'm unable to see any connection in those given information. How can I solve this logarithm?





logarithms





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 29 at 6:57









PJK

5,01811127




5,01811127









asked Jul 29 at 6:55









Steve

153




153




closed as off-topic by Henrik, John Ma, Xander Henderson, amWhy, user223391 Jul 31 at 1:08


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." – Henrik, John Ma, Xander Henderson, amWhy, Community
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Henrik, John Ma, Xander Henderson, amWhy, user223391 Jul 31 at 1:08


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." – Henrik, John Ma, Xander Henderson, amWhy, Community
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    I'm unable to see any connection in those given information What do you know about logarithms change of base?
    – dxiv
    Jul 29 at 6:58










  • @dxiv you mean this notation: $log_ab=frac1log_ba$?
    – Steve
    Jul 29 at 7:03










  • @Steve: That's not notation, that's a formula - that seems relevant in this case - but I would just use $log_a b=fraclog blog a$ (from which your formula follows).
    – Henrik
    Jul 29 at 7:07






  • 2




    @Steve And more generally $log _ab=frac log _cblog _ca.,$
    – dxiv
    Jul 29 at 7:09






  • 1




    Why don't you just solve for $a$ and substitute?
    – Miksu
    Jul 29 at 7:58












  • 1




    I'm unable to see any connection in those given information What do you know about logarithms change of base?
    – dxiv
    Jul 29 at 6:58










  • @dxiv you mean this notation: $log_ab=frac1log_ba$?
    – Steve
    Jul 29 at 7:03










  • @Steve: That's not notation, that's a formula - that seems relevant in this case - but I would just use $log_a b=fraclog blog a$ (from which your formula follows).
    – Henrik
    Jul 29 at 7:07






  • 2




    @Steve And more generally $log _ab=frac log _cblog _ca.,$
    – dxiv
    Jul 29 at 7:09






  • 1




    Why don't you just solve for $a$ and substitute?
    – Miksu
    Jul 29 at 7:58







1




1




I'm unable to see any connection in those given information What do you know about logarithms change of base?
– dxiv
Jul 29 at 6:58




I'm unable to see any connection in those given information What do you know about logarithms change of base?
– dxiv
Jul 29 at 6:58












@dxiv you mean this notation: $log_ab=frac1log_ba$?
– Steve
Jul 29 at 7:03




@dxiv you mean this notation: $log_ab=frac1log_ba$?
– Steve
Jul 29 at 7:03












@Steve: That's not notation, that's a formula - that seems relevant in this case - but I would just use $log_a b=fraclog blog a$ (from which your formula follows).
– Henrik
Jul 29 at 7:07




@Steve: That's not notation, that's a formula - that seems relevant in this case - but I would just use $log_a b=fraclog blog a$ (from which your formula follows).
– Henrik
Jul 29 at 7:07




2




2




@Steve And more generally $log _ab=frac log _cblog _ca.,$
– dxiv
Jul 29 at 7:09




@Steve And more generally $log _ab=frac log _cblog _ca.,$
– dxiv
Jul 29 at 7:09




1




1




Why don't you just solve for $a$ and substitute?
– Miksu
Jul 29 at 7:58




Why don't you just solve for $a$ and substitute?
– Miksu
Jul 29 at 7:58










1 Answer
1






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Given $log_27(a) = b$ then $a = 27^b$. Then:
$$sqrt[6]a = (27^b)^frac16 = (3^3 cdot frac16)^b = 3^fracb2$$
$$log_sqrt[6]ax = log_3^fracb2x = fraclog_3xlog_33^fracb2 = frac2blog_3x$$
Since $x = sqrt3$ then
$$frac2blog_3x = frac2bcdotfrac12 = frac1b$$






share|cite|improve this answer






























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    Given $log_27(a) = b$ then $a = 27^b$. Then:
    $$sqrt[6]a = (27^b)^frac16 = (3^3 cdot frac16)^b = 3^fracb2$$
    $$log_sqrt[6]ax = log_3^fracb2x = fraclog_3xlog_33^fracb2 = frac2blog_3x$$
    Since $x = sqrt3$ then
    $$frac2blog_3x = frac2bcdotfrac12 = frac1b$$






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      Given $log_27(a) = b$ then $a = 27^b$. Then:
      $$sqrt[6]a = (27^b)^frac16 = (3^3 cdot frac16)^b = 3^fracb2$$
      $$log_sqrt[6]ax = log_3^fracb2x = fraclog_3xlog_33^fracb2 = frac2blog_3x$$
      Since $x = sqrt3$ then
      $$frac2blog_3x = frac2bcdotfrac12 = frac1b$$






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Given $log_27(a) = b$ then $a = 27^b$. Then:
        $$sqrt[6]a = (27^b)^frac16 = (3^3 cdot frac16)^b = 3^fracb2$$
        $$log_sqrt[6]ax = log_3^fracb2x = fraclog_3xlog_33^fracb2 = frac2blog_3x$$
        Since $x = sqrt3$ then
        $$frac2blog_3x = frac2bcdotfrac12 = frac1b$$






        share|cite|improve this answer















        Given $log_27(a) = b$ then $a = 27^b$. Then:
        $$sqrt[6]a = (27^b)^frac16 = (3^3 cdot frac16)^b = 3^fracb2$$
        $$log_sqrt[6]ax = log_3^fracb2x = fraclog_3xlog_33^fracb2 = frac2blog_3x$$
        Since $x = sqrt3$ then
        $$frac2blog_3x = frac2bcdotfrac12 = frac1b$$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 2 at 7:30


























        answered Jul 29 at 8:02









        Giulio Scattolin

        15619




        15619












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