Middle School Log question












1














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



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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    If you have an acceptable answer, accept it by clicking the check mark on that answer.
    – Sean Roberson
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1














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



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







share|cite|improve this question




















  • 3




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














1












1








1







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



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







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




logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Rócherz

2,7362721




2,7362721










asked 1 hour ago









Toylatte

113




113








  • 3




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














  • 3




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








3




3




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




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










2 Answers
2






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oldest

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6














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
    1 hour ago










  • @Toylatte You are welcome!
    – ImNotTheGuy
    1 hour ago



















3














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



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





















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    2 Answers
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    active

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    2 Answers
    2






    active

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    active

    oldest

    votes






    active

    oldest

    votes









    6














    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
      1 hour ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      1 hour ago
















    6














    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
      1 hour ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      1 hour ago














    6












    6








    6






    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.
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    answered 1 hour ago









    ImNotTheGuy

    3825




    3825




    New contributor




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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
      1 hour ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      1 hour ago


















    • Thanks it helped
      – Toylatte
      1 hour ago










    • @Toylatte You are welcome!
      – ImNotTheGuy
      1 hour ago
















    Thanks it helped
    – Toylatte
    1 hour ago




    Thanks it helped
    – Toylatte
    1 hour ago












    @Toylatte You are welcome!
    – ImNotTheGuy
    1 hour ago




    @Toylatte You are welcome!
    – ImNotTheGuy
    1 hour ago











    3














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



    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


























      3














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



      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
























        3












        3








        3






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



        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 1 hour ago









        Siong Thye Goh

        98.8k1464116




        98.8k1464116






























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