Middle School Log question












2














$$left(frac 43right)^x=frac{8sqrt3}{9}$$



I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.










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    If you have an acceptable answer, accept it by clicking the check mark on that answer.
    – Sean Roberson
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2














$$left(frac 43right)^x=frac{8sqrt3}{9}$$



I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.










share|cite|improve this question




















  • 1




    If you have an acceptable answer, accept it by clicking the check mark on that answer.
    – Sean Roberson
    27 mins ago














2












2








2







$$left(frac 43right)^x=frac{8sqrt3}{9}$$



I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.










share|cite|improve this question















$$left(frac 43right)^x=frac{8sqrt3}{9}$$



I've been trying to solve this question. I know that the answer is $3/2$ by guess and check, but I want to know how to do it by actually solving it.







logarithms






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edited 14 mins ago









Rócherz

2,7362721




2,7362721










asked 47 mins ago









Toylatte

133




133








  • 1




    If you have an acceptable answer, accept it by clicking the check mark on that answer.
    – Sean Roberson
    27 mins ago














  • 1




    If you have an acceptable answer, accept it by clicking the check mark on that answer.
    – Sean Roberson
    27 mins ago








1




1




If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
27 mins ago




If you have an acceptable answer, accept it by clicking the check mark on that answer.
– Sean Roberson
27 mins ago










2 Answers
2






active

oldest

votes


















4














I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
$$
left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
$$

Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!



Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.






share|cite|improve this answer








New contributor




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


















  • Thanks it helped
    – Toylatte
    39 mins ago










  • @Toylatte You are welcome!
    – ImNotTheGuy
    39 mins ago



















2














$$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$



Note that the function $f(x)=left( frac43right)^x$ is an increasing function.



Hence $x=1.5$.



Alternatively, taking logarithm on both sides.






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
    $$
    left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
    $$

    Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!



    Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.






    share|cite|improve this answer








    New contributor




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


















    • Thanks it helped
      – Toylatte
      39 mins ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      39 mins ago
















    4














    I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
    $$
    left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
    $$

    Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!



    Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.






    share|cite|improve this answer








    New contributor




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


















    • Thanks it helped
      – Toylatte
      39 mins ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      39 mins ago














    4












    4








    4






    I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
    $$
    left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
    $$

    Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!



    Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.






    share|cite|improve this answer








    New contributor




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









    I think that perhaps the best way of tackling this problem is not through logarithms, but rather by raising both sides of your equation to another power:
    $$
    left(frac{4}{3}right)^x = frac{8sqrt{3}}{9} quad Rightarrow quad left(frac{4}{3}right)^{2x} = left(frac{8sqrt{3}}{9}right)^2
    $$

    Now things are looking up! The right hand side (RHS) is now equal to $64/27$, which we can much more readily identify as $4^3/3^3$. But if I need to cube $4/3$ then what value is $x$? Well, $2x=3$ implies $x=3/2$!



    Generally speaking, logarithms with non-integer bases are hard to work with; when possible, instead just continue to work with exponents.







    share|cite|improve this answer








    New contributor




    ImNotTheGuy 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 answer



    share|cite|improve this answer






    New contributor




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









    answered 42 mins ago









    ImNotTheGuy

    3364




    3364




    New contributor




    ImNotTheGuy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    New contributor





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






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












    • Thanks it helped
      – Toylatte
      39 mins ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      39 mins ago


















    • Thanks it helped
      – Toylatte
      39 mins ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      39 mins ago
















    Thanks it helped
    – Toylatte
    39 mins ago




    Thanks it helped
    – Toylatte
    39 mins ago












    @Toylatte You are welcome!
    – ImNotTheGuy
    39 mins ago




    @Toylatte You are welcome!
    – ImNotTheGuy
    39 mins ago











    2














    $$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$



    Note that the function $f(x)=left( frac43right)^x$ is an increasing function.



    Hence $x=1.5$.



    Alternatively, taking logarithm on both sides.






    share|cite|improve this answer


























      2














      $$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$



      Note that the function $f(x)=left( frac43right)^x$ is an increasing function.



      Hence $x=1.5$.



      Alternatively, taking logarithm on both sides.






      share|cite|improve this answer
























        2












        2








        2






        $$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$



        Note that the function $f(x)=left( frac43right)^x$ is an increasing function.



        Hence $x=1.5$.



        Alternatively, taking logarithm on both sides.






        share|cite|improve this answer












        $$frac{8sqrt3}{9}=frac{2^3cdot 3^{0.5}}{3^2}=frac{2^3}{3^{1.5}}=left(frac{2^2}{3}right)^{1.5}$$



        Note that the function $f(x)=left( frac43right)^x$ is an increasing function.



        Hence $x=1.5$.



        Alternatively, taking logarithm on both sides.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 42 mins ago









        Siong Thye Goh

        98.8k1464116




        98.8k1464116






























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