Help with solving an in-equation with a variable as exponent












0














I need to decide on an integer value for $n$ in the following in-equation:



$left(1-dfrac{1}{2^{64}}right)^n leq dfrac{1}{2}$



What I have tried:



The expression above reminds me of the known limit:



$limlimits_{n to infty}left(1 - dfrac{1}{n}right)^n = dfrac{1}{e} $



So, my intuition is to just determine $n = 2^{64}$ and to evaluate the entire left-hand-side expression as approaching to $dfrac{1}{e}$ and since that is less than $dfrac{1}{2}$ then it "feels" true



However, I feel this is really not a the way of solving this equation and there can probably be a better way to do it and obviously get a better value for $n$.



Thanks in advance






limits inequality







share|cite|improve this question




















  • 3




    May be take $log_2$ on both sides.
    – Yadati Kiran
    Nov 22 at 21:54
















0














I need to decide on an integer value for $n$ in the following in-equation:



$left(1-dfrac{1}{2^{64}}right)^n leq dfrac{1}{2}$



What I have tried:



The expression above reminds me of the known limit:



$limlimits_{n to infty}left(1 - dfrac{1}{n}right)^n = dfrac{1}{e} $



So, my intuition is to just determine $n = 2^{64}$ and to evaluate the entire left-hand-side expression as approaching to $dfrac{1}{e}$ and since that is less than $dfrac{1}{2}$ then it "feels" true



However, I feel this is really not a the way of solving this equation and there can probably be a better way to do it and obviously get a better value for $n$.



Thanks in advance






limits inequality







share|cite|improve this question




















  • 3




    May be take $log_2$ on both sides.
    – Yadati Kiran
    Nov 22 at 21:54














0












0








0


1





I need to decide on an integer value for $n$ in the following in-equation:



$left(1-dfrac{1}{2^{64}}right)^n leq dfrac{1}{2}$



What I have tried:



The expression above reminds me of the known limit:



$limlimits_{n to infty}left(1 - dfrac{1}{n}right)^n = dfrac{1}{e} $



So, my intuition is to just determine $n = 2^{64}$ and to evaluate the entire left-hand-side expression as approaching to $dfrac{1}{e}$ and since that is less than $dfrac{1}{2}$ then it "feels" true



However, I feel this is really not a the way of solving this equation and there can probably be a better way to do it and obviously get a better value for $n$.



Thanks in advance






limits inequality







share|cite|improve this question















I need to decide on an integer value for $n$ in the following in-equation:



$left(1-dfrac{1}{2^{64}}right)^n leq dfrac{1}{2}$



What I have tried:



The expression above reminds me of the known limit:



$limlimits_{n to infty}left(1 - dfrac{1}{n}right)^n = dfrac{1}{e} $



So, my intuition is to just determine $n = 2^{64}$ and to evaluate the entire left-hand-side expression as approaching to $dfrac{1}{e}$ and since that is less than $dfrac{1}{2}$ then it "feels" true



However, I feel this is really not a the way of solving this equation and there can probably be a better way to do it and obviously get a better value for $n$.



Thanks in advance






limits inequality




limits inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 22:00









Yadati Kiran

1,694519




1,694519










asked Nov 22 at 21:51









shaqed

28537




28537








  • 3




    May be take $log_2$ on both sides.
    – Yadati Kiran
    Nov 22 at 21:54














  • 3




    May be take $log_2$ on both sides.
    – Yadati Kiran
    Nov 22 at 21:54








3




3




May be take $log_2$ on both sides.
– Yadati Kiran
Nov 22 at 21:54




May be take $log_2$ on both sides.
– Yadati Kiran
Nov 22 at 21:54










2 Answers
2






active

oldest

votes


















2














Since $log$ function is increasing, we have that



$$left(1-frac{1}{2^{64}}right)^n leq frac{1}{2} iff nlogleft(1-frac{1}{2^{64}}right) leq log frac{1}{2} iff ngefrac{log frac{1}{2}}{logleft(1-frac{1}{2^{64}}right)}approx1.28cdot 10^{19}$$






share|cite|improve this answer































    2














    If OP wants some $n$ to satisfy the inequality they can proceed as follows. From the inequality
    $$
    e^xgeq 1+x
    $$
    for all $x$ we have that
    $$
    left(1-frac{1}{2^{64}}right)^nleq e^{-n/2^{64}}
    $$
    and
    $$
    e^{-n/2^{64}}leq frac{1}{2}
    $$
    if
    $$
    ngeq 2^{64}times log 2geq 2^{32}
    $$
    Hence $n=2^{32}$ works for example.






    share|cite|improve this answer























    • I couldn't quite understand the first step, could you please elaborate? where is $2^{64}$ being substituted exactly? why is $n$ in the exponent of $e$ ?
      – shaqed
      Nov 23 at 13:10











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    Since $log$ function is increasing, we have that



    $$left(1-frac{1}{2^{64}}right)^n leq frac{1}{2} iff nlogleft(1-frac{1}{2^{64}}right) leq log frac{1}{2} iff ngefrac{log frac{1}{2}}{logleft(1-frac{1}{2^{64}}right)}approx1.28cdot 10^{19}$$






    share|cite|improve this answer




























      2














      Since $log$ function is increasing, we have that



      $$left(1-frac{1}{2^{64}}right)^n leq frac{1}{2} iff nlogleft(1-frac{1}{2^{64}}right) leq log frac{1}{2} iff ngefrac{log frac{1}{2}}{logleft(1-frac{1}{2^{64}}right)}approx1.28cdot 10^{19}$$






      share|cite|improve this answer


























        2












        2








        2






        Since $log$ function is increasing, we have that



        $$left(1-frac{1}{2^{64}}right)^n leq frac{1}{2} iff nlogleft(1-frac{1}{2^{64}}right) leq log frac{1}{2} iff ngefrac{log frac{1}{2}}{logleft(1-frac{1}{2^{64}}right)}approx1.28cdot 10^{19}$$






        share|cite|improve this answer














        Since $log$ function is increasing, we have that



        $$left(1-frac{1}{2^{64}}right)^n leq frac{1}{2} iff nlogleft(1-frac{1}{2^{64}}right) leq log frac{1}{2} iff ngefrac{log frac{1}{2}}{logleft(1-frac{1}{2^{64}}right)}approx1.28cdot 10^{19}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 22 at 21:59

























        answered Nov 22 at 21:54









        gimusi

        1




        1























            2














            If OP wants some $n$ to satisfy the inequality they can proceed as follows. From the inequality
            $$
            e^xgeq 1+x
            $$
            for all $x$ we have that
            $$
            left(1-frac{1}{2^{64}}right)^nleq e^{-n/2^{64}}
            $$
            and
            $$
            e^{-n/2^{64}}leq frac{1}{2}
            $$
            if
            $$
            ngeq 2^{64}times log 2geq 2^{32}
            $$
            Hence $n=2^{32}$ works for example.






            share|cite|improve this answer























            • I couldn't quite understand the first step, could you please elaborate? where is $2^{64}$ being substituted exactly? why is $n$ in the exponent of $e$ ?
              – shaqed
              Nov 23 at 13:10
















            2














            If OP wants some $n$ to satisfy the inequality they can proceed as follows. From the inequality
            $$
            e^xgeq 1+x
            $$
            for all $x$ we have that
            $$
            left(1-frac{1}{2^{64}}right)^nleq e^{-n/2^{64}}
            $$
            and
            $$
            e^{-n/2^{64}}leq frac{1}{2}
            $$
            if
            $$
            ngeq 2^{64}times log 2geq 2^{32}
            $$
            Hence $n=2^{32}$ works for example.






            share|cite|improve this answer























            • I couldn't quite understand the first step, could you please elaborate? where is $2^{64}$ being substituted exactly? why is $n$ in the exponent of $e$ ?
              – shaqed
              Nov 23 at 13:10














            2












            2








            2






            If OP wants some $n$ to satisfy the inequality they can proceed as follows. From the inequality
            $$
            e^xgeq 1+x
            $$
            for all $x$ we have that
            $$
            left(1-frac{1}{2^{64}}right)^nleq e^{-n/2^{64}}
            $$
            and
            $$
            e^{-n/2^{64}}leq frac{1}{2}
            $$
            if
            $$
            ngeq 2^{64}times log 2geq 2^{32}
            $$
            Hence $n=2^{32}$ works for example.






            share|cite|improve this answer














            If OP wants some $n$ to satisfy the inequality they can proceed as follows. From the inequality
            $$
            e^xgeq 1+x
            $$
            for all $x$ we have that
            $$
            left(1-frac{1}{2^{64}}right)^nleq e^{-n/2^{64}}
            $$
            and
            $$
            e^{-n/2^{64}}leq frac{1}{2}
            $$
            if
            $$
            ngeq 2^{64}times log 2geq 2^{32}
            $$
            Hence $n=2^{32}$ works for example.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 22 at 22:02

























            answered Nov 22 at 21:57









            Foobaz John

            20.8k41250




            20.8k41250












            • I couldn't quite understand the first step, could you please elaborate? where is $2^{64}$ being substituted exactly? why is $n$ in the exponent of $e$ ?
              – shaqed
              Nov 23 at 13:10


















            • I couldn't quite understand the first step, could you please elaborate? where is $2^{64}$ being substituted exactly? why is $n$ in the exponent of $e$ ?
              – shaqed
              Nov 23 at 13:10
















            I couldn't quite understand the first step, could you please elaborate? where is $2^{64}$ being substituted exactly? why is $n$ in the exponent of $e$ ?
            – shaqed
            Nov 23 at 13:10




            I couldn't quite understand the first step, could you please elaborate? where is $2^{64}$ being substituted exactly? why is $n$ in the exponent of $e$ ?
            – shaqed
            Nov 23 at 13:10


















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