Estimate the value of $log_{20}3$












2












$begingroup$



Estimate the value of $log_{20}3$




My Attempt
$$
y=log_{20}3=frac{1}{log_{3}20}=frac{log_{c}3}{log_c20}=frac{log_c3}{log_c5+log_c4}\
=frac{1}{frac{log_c5}{log_c3}+frac{log_c4}{log_c3}}=frac{1}{log_35+log_34}\
x=log_35implies3^x=5implies x<2;&;x>1\
z=log_34implies3^z=4implies z<2;&;z>1\
x+z<4;&;x+z>2\
yinBig(frac{1}{4},frac{1}{2}Big)
$$

My reference gives the solution $Big(frac{1}{3},frac{1}{2}Big)$, it seems $frac{1}{4}$ is not the lowest limit of $log_{20}3$, what's the easiest way to calculate it ?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
    $endgroup$
    – Satish Ramanathan
    Jan 21 at 8:03






  • 1




    $begingroup$
    Have you copied the problem correctly?
    $endgroup$
    – Satish Ramanathan
    Jan 21 at 8:04






  • 7




    $begingroup$
    FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 21 at 8:05






  • 2




    $begingroup$
    The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
    $endgroup$
    – Yves Daoust
    Jan 21 at 8:14






  • 1




    $begingroup$
    Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
    $endgroup$
    – zwim
    Jan 21 at 8:58


















2












$begingroup$



Estimate the value of $log_{20}3$




My Attempt
$$
y=log_{20}3=frac{1}{log_{3}20}=frac{log_{c}3}{log_c20}=frac{log_c3}{log_c5+log_c4}\
=frac{1}{frac{log_c5}{log_c3}+frac{log_c4}{log_c3}}=frac{1}{log_35+log_34}\
x=log_35implies3^x=5implies x<2;&;x>1\
z=log_34implies3^z=4implies z<2;&;z>1\
x+z<4;&;x+z>2\
yinBig(frac{1}{4},frac{1}{2}Big)
$$

My reference gives the solution $Big(frac{1}{3},frac{1}{2}Big)$, it seems $frac{1}{4}$ is not the lowest limit of $log_{20}3$, what's the easiest way to calculate it ?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
    $endgroup$
    – Satish Ramanathan
    Jan 21 at 8:03






  • 1




    $begingroup$
    Have you copied the problem correctly?
    $endgroup$
    – Satish Ramanathan
    Jan 21 at 8:04






  • 7




    $begingroup$
    FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 21 at 8:05






  • 2




    $begingroup$
    The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
    $endgroup$
    – Yves Daoust
    Jan 21 at 8:14






  • 1




    $begingroup$
    Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
    $endgroup$
    – zwim
    Jan 21 at 8:58
















2












2








2


1



$begingroup$



Estimate the value of $log_{20}3$




My Attempt
$$
y=log_{20}3=frac{1}{log_{3}20}=frac{log_{c}3}{log_c20}=frac{log_c3}{log_c5+log_c4}\
=frac{1}{frac{log_c5}{log_c3}+frac{log_c4}{log_c3}}=frac{1}{log_35+log_34}\
x=log_35implies3^x=5implies x<2;&;x>1\
z=log_34implies3^z=4implies z<2;&;z>1\
x+z<4;&;x+z>2\
yinBig(frac{1}{4},frac{1}{2}Big)
$$

My reference gives the solution $Big(frac{1}{3},frac{1}{2}Big)$, it seems $frac{1}{4}$ is not the lowest limit of $log_{20}3$, what's the easiest way to calculate it ?










share|cite|improve this question











$endgroup$





Estimate the value of $log_{20}3$




My Attempt
$$
y=log_{20}3=frac{1}{log_{3}20}=frac{log_{c}3}{log_c20}=frac{log_c3}{log_c5+log_c4}\
=frac{1}{frac{log_c5}{log_c3}+frac{log_c4}{log_c3}}=frac{1}{log_35+log_34}\
x=log_35implies3^x=5implies x<2;&;x>1\
z=log_34implies3^z=4implies z<2;&;z>1\
x+z<4;&;x+z>2\
yinBig(frac{1}{4},frac{1}{2}Big)
$$

My reference gives the solution $Big(frac{1}{3},frac{1}{2}Big)$, it seems $frac{1}{4}$ is not the lowest limit of $log_{20}3$, what's the easiest way to calculate it ?







number-theory logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 9:42







ss1729

















asked Jan 21 at 7:54









ss1729ss1729

2,00111024




2,00111024








  • 1




    $begingroup$
    y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
    $endgroup$
    – Satish Ramanathan
    Jan 21 at 8:03






  • 1




    $begingroup$
    Have you copied the problem correctly?
    $endgroup$
    – Satish Ramanathan
    Jan 21 at 8:04






  • 7




    $begingroup$
    FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 21 at 8:05






  • 2




    $begingroup$
    The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
    $endgroup$
    – Yves Daoust
    Jan 21 at 8:14






  • 1




    $begingroup$
    Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
    $endgroup$
    – zwim
    Jan 21 at 8:58
















  • 1




    $begingroup$
    y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
    $endgroup$
    – Satish Ramanathan
    Jan 21 at 8:03






  • 1




    $begingroup$
    Have you copied the problem correctly?
    $endgroup$
    – Satish Ramanathan
    Jan 21 at 8:04






  • 7




    $begingroup$
    FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 21 at 8:05






  • 2




    $begingroup$
    The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
    $endgroup$
    – Yves Daoust
    Jan 21 at 8:14






  • 1




    $begingroup$
    Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
    $endgroup$
    – zwim
    Jan 21 at 8:58










1




1




$begingroup$
y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:03




$begingroup$
y= constant has a range which includes just that constant. there is no x anywhere and I wonder how you can find a range?. Am I missing something?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:03




1




1




$begingroup$
Have you copied the problem correctly?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:04




$begingroup$
Have you copied the problem correctly?
$endgroup$
– Satish Ramanathan
Jan 21 at 8:04




7




7




$begingroup$
FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 21 at 8:05




$begingroup$
FWIW, I think the problem should be restated as: Estimate $log_{20}{3}$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 21 at 8:05




2




2




$begingroup$
The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
$endgroup$
– Yves Daoust
Jan 21 at 8:14




$begingroup$
The range of a number makes little sense. You can find arbitrarily close approximations, there is no unique answer.
$endgroup$
– Yves Daoust
Jan 21 at 8:14




1




1




$begingroup$
Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
$endgroup$
– zwim
Jan 21 at 8:58






$begingroup$
Btw, in French we would use encadrement for a double inequality involving a quantity and its lower and upper bounds. Is there an equivalent single word in English ? I suspect this is why range was used.
$endgroup$
– zwim
Jan 21 at 8:58












2 Answers
2






active

oldest

votes


















7












$begingroup$

Estimation of $log_{20}(3)$



Using $$9<20<27.$$



$$log_{3}(9)<log_{3}(20)<log_{3}(27)$$
So $$2<log_{3}(20)<3$$



So $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$






share|cite|improve this answer









$endgroup$





















    9












    $begingroup$

    Write the sequence of powers of $3$ and $20$:



    $$1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,cdots$$



    $$1,20,400,8000,160000,3200000,64000000,cdots$$



    This gives you many rational upper and lower bounds:



    $$3^2<20^1implies 2log3<1log20implieslog_{20}3<frac12$$



    $$3^5<20^2implies 5log3<2log20implieslog_{20}3<frac25$$



    $$3^8<20^3implies 8log3<3log20implieslog_{20}3<frac38$$



    $$3^{10}<20^4implies 10log3<4log20implieslog_{20}3<frac25$$



    $$cdots$$



    $$3^3>20^1implies 3log3>1log20implieslog_{20}3>frac13$$



    $$3^6>20^2implies 6log3>2log20implieslog_{20}3>frac13$$



    $$3^9>20^3implies 9log3>3log20implieslog_{20}3>frac13$$



    $$3^{11}>20^4implies 11log3>4log20implieslog_{20}3>frac4{11}$$



    $$cdots$$



    By hand, you could reasonably establish



    $$frac4{11}<log_{20}3<frac7{19}$$



    $$0.3636364<0.3667258<0.3684211$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Nice manipulation !
      $endgroup$
      – Claude Leibovici
      Jan 21 at 9:06










    • $begingroup$
      @ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
      $endgroup$
      – Yves Daoust
      Jan 21 at 9:08













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

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    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    Estimation of $log_{20}(3)$



    Using $$9<20<27.$$



    $$log_{3}(9)<log_{3}(20)<log_{3}(27)$$
    So $$2<log_{3}(20)<3$$



    So $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$






    share|cite|improve this answer









    $endgroup$


















      7












      $begingroup$

      Estimation of $log_{20}(3)$



      Using $$9<20<27.$$



      $$log_{3}(9)<log_{3}(20)<log_{3}(27)$$
      So $$2<log_{3}(20)<3$$



      So $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$






      share|cite|improve this answer









      $endgroup$
















        7












        7








        7





        $begingroup$

        Estimation of $log_{20}(3)$



        Using $$9<20<27.$$



        $$log_{3}(9)<log_{3}(20)<log_{3}(27)$$
        So $$2<log_{3}(20)<3$$



        So $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$






        share|cite|improve this answer









        $endgroup$



        Estimation of $log_{20}(3)$



        Using $$9<20<27.$$



        $$log_{3}(9)<log_{3}(20)<log_{3}(27)$$
        So $$2<log_{3}(20)<3$$



        So $$frac{1}{3}<log_{20}(3)<frac{1}{2}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 21 at 8:08









        DXTDXT

        5,9742732




        5,9742732























            9












            $begingroup$

            Write the sequence of powers of $3$ and $20$:



            $$1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,cdots$$



            $$1,20,400,8000,160000,3200000,64000000,cdots$$



            This gives you many rational upper and lower bounds:



            $$3^2<20^1implies 2log3<1log20implieslog_{20}3<frac12$$



            $$3^5<20^2implies 5log3<2log20implieslog_{20}3<frac25$$



            $$3^8<20^3implies 8log3<3log20implieslog_{20}3<frac38$$



            $$3^{10}<20^4implies 10log3<4log20implieslog_{20}3<frac25$$



            $$cdots$$



            $$3^3>20^1implies 3log3>1log20implieslog_{20}3>frac13$$



            $$3^6>20^2implies 6log3>2log20implieslog_{20}3>frac13$$



            $$3^9>20^3implies 9log3>3log20implieslog_{20}3>frac13$$



            $$3^{11}>20^4implies 11log3>4log20implieslog_{20}3>frac4{11}$$



            $$cdots$$



            By hand, you could reasonably establish



            $$frac4{11}<log_{20}3<frac7{19}$$



            $$0.3636364<0.3667258<0.3684211$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Nice manipulation !
              $endgroup$
              – Claude Leibovici
              Jan 21 at 9:06










            • $begingroup$
              @ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
              $endgroup$
              – Yves Daoust
              Jan 21 at 9:08


















            9












            $begingroup$

            Write the sequence of powers of $3$ and $20$:



            $$1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,cdots$$



            $$1,20,400,8000,160000,3200000,64000000,cdots$$



            This gives you many rational upper and lower bounds:



            $$3^2<20^1implies 2log3<1log20implieslog_{20}3<frac12$$



            $$3^5<20^2implies 5log3<2log20implieslog_{20}3<frac25$$



            $$3^8<20^3implies 8log3<3log20implieslog_{20}3<frac38$$



            $$3^{10}<20^4implies 10log3<4log20implieslog_{20}3<frac25$$



            $$cdots$$



            $$3^3>20^1implies 3log3>1log20implieslog_{20}3>frac13$$



            $$3^6>20^2implies 6log3>2log20implieslog_{20}3>frac13$$



            $$3^9>20^3implies 9log3>3log20implieslog_{20}3>frac13$$



            $$3^{11}>20^4implies 11log3>4log20implieslog_{20}3>frac4{11}$$



            $$cdots$$



            By hand, you could reasonably establish



            $$frac4{11}<log_{20}3<frac7{19}$$



            $$0.3636364<0.3667258<0.3684211$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Nice manipulation !
              $endgroup$
              – Claude Leibovici
              Jan 21 at 9:06










            • $begingroup$
              @ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
              $endgroup$
              – Yves Daoust
              Jan 21 at 9:08
















            9












            9








            9





            $begingroup$

            Write the sequence of powers of $3$ and $20$:



            $$1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,cdots$$



            $$1,20,400,8000,160000,3200000,64000000,cdots$$



            This gives you many rational upper and lower bounds:



            $$3^2<20^1implies 2log3<1log20implieslog_{20}3<frac12$$



            $$3^5<20^2implies 5log3<2log20implieslog_{20}3<frac25$$



            $$3^8<20^3implies 8log3<3log20implieslog_{20}3<frac38$$



            $$3^{10}<20^4implies 10log3<4log20implieslog_{20}3<frac25$$



            $$cdots$$



            $$3^3>20^1implies 3log3>1log20implieslog_{20}3>frac13$$



            $$3^6>20^2implies 6log3>2log20implieslog_{20}3>frac13$$



            $$3^9>20^3implies 9log3>3log20implieslog_{20}3>frac13$$



            $$3^{11}>20^4implies 11log3>4log20implieslog_{20}3>frac4{11}$$



            $$cdots$$



            By hand, you could reasonably establish



            $$frac4{11}<log_{20}3<frac7{19}$$



            $$0.3636364<0.3667258<0.3684211$$






            share|cite|improve this answer











            $endgroup$



            Write the sequence of powers of $3$ and $20$:



            $$1,3,9,27,81,243,729,2187,6561,19683,59049,177147,531441,1594323,4782969,14348907,cdots$$



            $$1,20,400,8000,160000,3200000,64000000,cdots$$



            This gives you many rational upper and lower bounds:



            $$3^2<20^1implies 2log3<1log20implieslog_{20}3<frac12$$



            $$3^5<20^2implies 5log3<2log20implieslog_{20}3<frac25$$



            $$3^8<20^3implies 8log3<3log20implieslog_{20}3<frac38$$



            $$3^{10}<20^4implies 10log3<4log20implieslog_{20}3<frac25$$



            $$cdots$$



            $$3^3>20^1implies 3log3>1log20implieslog_{20}3>frac13$$



            $$3^6>20^2implies 6log3>2log20implieslog_{20}3>frac13$$



            $$3^9>20^3implies 9log3>3log20implieslog_{20}3>frac13$$



            $$3^{11}>20^4implies 11log3>4log20implieslog_{20}3>frac4{11}$$



            $$cdots$$



            By hand, you could reasonably establish



            $$frac4{11}<log_{20}3<frac7{19}$$



            $$0.3636364<0.3667258<0.3684211$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 21 at 9:41

























            answered Jan 21 at 8:31









            Yves DaoustYves Daoust

            129k676227




            129k676227












            • $begingroup$
              Nice manipulation !
              $endgroup$
              – Claude Leibovici
              Jan 21 at 9:06










            • $begingroup$
              @ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
              $endgroup$
              – Yves Daoust
              Jan 21 at 9:08




















            • $begingroup$
              Nice manipulation !
              $endgroup$
              – Claude Leibovici
              Jan 21 at 9:06










            • $begingroup$
              @ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
              $endgroup$
              – Yves Daoust
              Jan 21 at 9:08


















            $begingroup$
            Nice manipulation !
            $endgroup$
            – Claude Leibovici
            Jan 21 at 9:06




            $begingroup$
            Nice manipulation !
            $endgroup$
            – Claude Leibovici
            Jan 21 at 9:06












            $begingroup$
            @ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
            $endgroup$
            – Yves Daoust
            Jan 21 at 9:08






            $begingroup$
            @ClaudeLeibovici: hi Claude. I didn't mention it, but using continuous fraction approximations, you can quickly find good bracketings by rationals (such as $(11/30,205/559)$), then prove them by the above method - by hand is a little tedious ;-). This spares many steps.
            $endgroup$
            – Yves Daoust
            Jan 21 at 9:08




















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