Middle School Log question












4














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



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
    4 hours ago
















4














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



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




















  • 5




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














4












4








4







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



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






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edited 4 hours ago









Rócherz

2,7612721




2,7612721










asked 4 hours ago









Toylatte

293




293








  • 5




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














  • 5




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








5




5




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




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










4 Answers
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13














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








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  • @Toylatte You are welcome!
    – ImNotTheGuy
    4 hours ago



















6














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



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





























    0














    Since you have tagged the question as a logarithmic question, let us try to approach it that way. Using the basic properties of log:
    $$
    (frac{4}{3})^x = frac{8 sqrt{3}}{9} \
    implies x log{4} - x log{3} = log{8 sqrt{3}} - log{9} \
    implies x log{4} - x log{3} = log{(2 times 4)} + log{3^{0.5}} - log{3^2} \
    implies log{4^x} - log{3^x} = log{4^{0.5}} + log{4^1} + log{3^{0.5}} - log{3^2} \
    implies log{4^x} - log{3^x} = log{4^{1.5}} - log{3^{-0.5}} - log{3^2} \
    implies log{4^x} - log{3^x} = log{4^{1.5}} - log{3^{1.5}} \
    implies log{(frac{4}{3})^x} = log{(frac{4}{3})^{1.5}}
    implies x log{frac{4}{3}} = 1.5 log{frac{4}{3}} \
    implies x = 1.5 = frac{3}{2} space square
    $$






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      0














      I'm not sure how deep you want to go into this with the rules of logarithms, but the simple answer is that



      $$
      log_b (x^n) = nlog_b (x).
      $$



      Thus, if you take the logarithm (say base 10) of both sides, then you have,



      $$
      log_{10} (4/3)^x = log_{10} (8sqrt 3/9)
      $$
      This leads to,
      $$
      x log_{10} (4/3) = log_{10} (8sqrt 3/9).
      $$



      If you divide both sides by $log_{10}(4/3)$ you isolate x and find your answer of 3/2.






      share|cite|improve this answer








      New contributor




      Thomas TJ Checkley is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        4 Answers
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        4 Answers
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        active

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        13














        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.


















        • @Toylatte You are welcome!
          – ImNotTheGuy
          4 hours ago
















        13














        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.


















        • @Toylatte You are welcome!
          – ImNotTheGuy
          4 hours ago














        13












        13








        13






        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.
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        share|cite|improve this answer



        share|cite|improve this answer






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        answered 4 hours ago









        ImNotTheGuy

        4066




        4066




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        • @Toylatte You are welcome!
          – ImNotTheGuy
          4 hours ago


















        • @Toylatte You are welcome!
          – ImNotTheGuy
          4 hours ago
















        @Toylatte You are welcome!
        – ImNotTheGuy
        4 hours ago




        @Toylatte You are welcome!
        – ImNotTheGuy
        4 hours ago











        6














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



        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


























          6














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



          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
























            6












            6








            6






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



            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 4 hours ago









            Siong Thye Goh

            98.8k1464116




            98.8k1464116























                0














                Since you have tagged the question as a logarithmic question, let us try to approach it that way. Using the basic properties of log:
                $$
                (frac{4}{3})^x = frac{8 sqrt{3}}{9} \
                implies x log{4} - x log{3} = log{8 sqrt{3}} - log{9} \
                implies x log{4} - x log{3} = log{(2 times 4)} + log{3^{0.5}} - log{3^2} \
                implies log{4^x} - log{3^x} = log{4^{0.5}} + log{4^1} + log{3^{0.5}} - log{3^2} \
                implies log{4^x} - log{3^x} = log{4^{1.5}} - log{3^{-0.5}} - log{3^2} \
                implies log{4^x} - log{3^x} = log{4^{1.5}} - log{3^{1.5}} \
                implies log{(frac{4}{3})^x} = log{(frac{4}{3})^{1.5}}
                implies x log{frac{4}{3}} = 1.5 log{frac{4}{3}} \
                implies x = 1.5 = frac{3}{2} space square
                $$






                share|cite|improve this answer








                New contributor




                Pratik K. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  0














                  Since you have tagged the question as a logarithmic question, let us try to approach it that way. Using the basic properties of log:
                  $$
                  (frac{4}{3})^x = frac{8 sqrt{3}}{9} \
                  implies x log{4} - x log{3} = log{8 sqrt{3}} - log{9} \
                  implies x log{4} - x log{3} = log{(2 times 4)} + log{3^{0.5}} - log{3^2} \
                  implies log{4^x} - log{3^x} = log{4^{0.5}} + log{4^1} + log{3^{0.5}} - log{3^2} \
                  implies log{4^x} - log{3^x} = log{4^{1.5}} - log{3^{-0.5}} - log{3^2} \
                  implies log{4^x} - log{3^x} = log{4^{1.5}} - log{3^{1.5}} \
                  implies log{(frac{4}{3})^x} = log{(frac{4}{3})^{1.5}}
                  implies x log{frac{4}{3}} = 1.5 log{frac{4}{3}} \
                  implies x = 1.5 = frac{3}{2} space square
                  $$






                  share|cite|improve this answer








                  New contributor




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





















                    0












                    0








                    0






                    Since you have tagged the question as a logarithmic question, let us try to approach it that way. Using the basic properties of log:
                    $$
                    (frac{4}{3})^x = frac{8 sqrt{3}}{9} \
                    implies x log{4} - x log{3} = log{8 sqrt{3}} - log{9} \
                    implies x log{4} - x log{3} = log{(2 times 4)} + log{3^{0.5}} - log{3^2} \
                    implies log{4^x} - log{3^x} = log{4^{0.5}} + log{4^1} + log{3^{0.5}} - log{3^2} \
                    implies log{4^x} - log{3^x} = log{4^{1.5}} - log{3^{-0.5}} - log{3^2} \
                    implies log{4^x} - log{3^x} = log{4^{1.5}} - log{3^{1.5}} \
                    implies log{(frac{4}{3})^x} = log{(frac{4}{3})^{1.5}}
                    implies x log{frac{4}{3}} = 1.5 log{frac{4}{3}} \
                    implies x = 1.5 = frac{3}{2} space square
                    $$






                    share|cite|improve this answer








                    New contributor




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









                    Since you have tagged the question as a logarithmic question, let us try to approach it that way. Using the basic properties of log:
                    $$
                    (frac{4}{3})^x = frac{8 sqrt{3}}{9} \
                    implies x log{4} - x log{3} = log{8 sqrt{3}} - log{9} \
                    implies x log{4} - x log{3} = log{(2 times 4)} + log{3^{0.5}} - log{3^2} \
                    implies log{4^x} - log{3^x} = log{4^{0.5}} + log{4^1} + log{3^{0.5}} - log{3^2} \
                    implies log{4^x} - log{3^x} = log{4^{1.5}} - log{3^{-0.5}} - log{3^2} \
                    implies log{4^x} - log{3^x} = log{4^{1.5}} - log{3^{1.5}} \
                    implies log{(frac{4}{3})^x} = log{(frac{4}{3})^{1.5}}
                    implies x log{frac{4}{3}} = 1.5 log{frac{4}{3}} \
                    implies x = 1.5 = frac{3}{2} space square
                    $$







                    share|cite|improve this answer








                    New contributor




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                    share|cite|improve this answer



                    share|cite|improve this answer






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                    answered 43 mins ago









                    Pratik K.

                    1




                    1




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                    New contributor





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                        0














                        I'm not sure how deep you want to go into this with the rules of logarithms, but the simple answer is that



                        $$
                        log_b (x^n) = nlog_b (x).
                        $$



                        Thus, if you take the logarithm (say base 10) of both sides, then you have,



                        $$
                        log_{10} (4/3)^x = log_{10} (8sqrt 3/9)
                        $$
                        This leads to,
                        $$
                        x log_{10} (4/3) = log_{10} (8sqrt 3/9).
                        $$



                        If you divide both sides by $log_{10}(4/3)$ you isolate x and find your answer of 3/2.






                        share|cite|improve this answer








                        New contributor




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























                          0














                          I'm not sure how deep you want to go into this with the rules of logarithms, but the simple answer is that



                          $$
                          log_b (x^n) = nlog_b (x).
                          $$



                          Thus, if you take the logarithm (say base 10) of both sides, then you have,



                          $$
                          log_{10} (4/3)^x = log_{10} (8sqrt 3/9)
                          $$
                          This leads to,
                          $$
                          x log_{10} (4/3) = log_{10} (8sqrt 3/9).
                          $$



                          If you divide both sides by $log_{10}(4/3)$ you isolate x and find your answer of 3/2.






                          share|cite|improve this answer








                          New contributor




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





















                            0












                            0








                            0






                            I'm not sure how deep you want to go into this with the rules of logarithms, but the simple answer is that



                            $$
                            log_b (x^n) = nlog_b (x).
                            $$



                            Thus, if you take the logarithm (say base 10) of both sides, then you have,



                            $$
                            log_{10} (4/3)^x = log_{10} (8sqrt 3/9)
                            $$
                            This leads to,
                            $$
                            x log_{10} (4/3) = log_{10} (8sqrt 3/9).
                            $$



                            If you divide both sides by $log_{10}(4/3)$ you isolate x and find your answer of 3/2.






                            share|cite|improve this answer








                            New contributor




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









                            I'm not sure how deep you want to go into this with the rules of logarithms, but the simple answer is that



                            $$
                            log_b (x^n) = nlog_b (x).
                            $$



                            Thus, if you take the logarithm (say base 10) of both sides, then you have,



                            $$
                            log_{10} (4/3)^x = log_{10} (8sqrt 3/9)
                            $$
                            This leads to,
                            $$
                            x log_{10} (4/3) = log_{10} (8sqrt 3/9).
                            $$



                            If you divide both sides by $log_{10}(4/3)$ you isolate x and find your answer of 3/2.







                            share|cite|improve this answer








                            New contributor




                            Thomas TJ Checkley 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




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                            answered 26 mins ago









                            Thomas TJ Checkley

                            11




                            11




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                                Volksrepublik China

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                                Dieser Artikel befasst sich mit der Volksrepublik China und ist nicht zu verwechseln mit der Republik China (Taiwan). Siehe auch China (Begriffsklärung). Dieser Artikel wurde aufgrund von inhaltlichen Mängeln auf der Qualitätssicherungsseite der Redaktion Ostasien eingetragen. Dies geschieht, um die Qualität der Artikel aus dem Themengebiet Ostasien auf ein akzeptables Niveau zu bringen. Dabei werden Artikel gelöscht, die nicht signifikant verbessert werden können. Bitte hilf mit, die inhaltlichen Mängel dieses Artikels zu beseitigen, und beteilige dich bitte an der Diskussion ! 中华人民共和国 Zhōnghuá Rénmín Gònghéguó Volksrepublik China Flagge Emblem Amtssprache Chinesisch (Hochchinesisch) und anerkannte Regionalsprachen 1 Hauptstadt Peking Staatsform Republik Regierungssystem Sozialistisches Einparteiensystem Staatsoberhaupt Staatspräsident Xi Jinping Regierungschef Ministerpräsident Li Keqiang Fläc...
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                                How to test boost logger output in unit testing?

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                                up vote 1 down vote favorite My application uses BOOST_LOG_TRIVIAL to log messages. I thought it's good to check in test (gtest) that correct messages are written there in given scenarios. Is there a way to somehow access what was written there after code was called? Or does it first have to be mocked in some way? I have googled a lot for this and haven't found any instructions, so either I am asking wrong questions or no one thinks it should be done. c++ logging boost googletest boost-log share | improve this question edited yesterday ...
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                                Write to the output between two pipeline

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                                up vote 1 down vote favorite I'm trying to write to the output (echo) within two pipe to have a trace of what I'm doing but I have an error. Get-ChildItem $path -Recurse | Where-Object { $_.Name -match '.+?.log.(d{4})-(d{2})-(d{2})$' -and $_.LastWriteTime -lt $deleteDate } | Write-Output "deleting file" -PassThru | Remove-Item but I have this error : Write-Output : The input object cannot be bound to any parameters for the command either because the command does not take pipeline input or the input and its properties do not match any of the parameters that take pipeline input. I also tested in the Foreach-Object function Get-ChildItem $path -Recurse | Where-Object { $_.Name -match '.+?.log.(d{4})-(d{2})-(d{2})$' -and $_.LastWriteTime -lt $deleteDate } | Foreach-Object...
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