bounding to get Number of digits without logsFind the number of digits in $2^2^22$ without logarithms.Difference between sum of even positioned digits and sum of odd positioned digits in a number is equal to 1prove that there are no number such that $overlineabldots=a^2+b^2+ldots$Is there an algebraic number which has all possible combinations of numbers?Last digits of a power of 2arbitrarily long sequences without perfect powersDistribution of digits in powers of 2Properties of The Number 137Can we generate arbitary long prime-number sequences by appending digits to a start prime?Repeatedly multiplying a number by itself - what explains the period of final digits?Sum of digits of sum of digits of sum of digits of $7^7^7^7$

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bounding to get Number of digits without logs


Find the number of digits in $2^2^22$ without logarithms.Difference between sum of even positioned digits and sum of odd positioned digits in a number is equal to 1prove that there are no number such that $overlineabldots=a^2+b^2+ldots$Is there an algebraic number which has all possible combinations of numbers?Last digits of a power of 2arbitrarily long sequences without perfect powersDistribution of digits in powers of 2Properties of The Number 137Can we generate arbitary long prime-number sequences by appending digits to a start prime?Repeatedly multiplying a number by itself - what explains the period of final digits?Sum of digits of sum of digits of sum of digits of $7^7^7^7$













1












$begingroup$


I would like to prove that $10^168 le 2^561< 10^169$ (to get the number of digits of the power of two, if possible without the use of logarithms.



The idea came to me from this post



Using the fact that $5^3<2^7$ gives $10^168<2^560$.



Any advice to improve this bound and get the other are welcome.



Thanks






number-theory







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Using log it is between $10^168$ and $10^169$ though.
    $endgroup$
    – Hw Chu
    Mar 14 at 17:36










  • $begingroup$
    using wolfram $floorlog(2^561) +1= 169$
    $endgroup$
    – ahmed
    Mar 14 at 17:37







  • 2




    $begingroup$
    Which means that $10^169$ is the upper bound.
    $endgroup$
    – Hw Chu
    Mar 14 at 17:39










  • $begingroup$
    fixed thanks, the RHS inequality is apparently the trickier
    $endgroup$
    – ahmed
    Mar 14 at 17:41















1












$begingroup$


I would like to prove that $10^168 le 2^561< 10^169$ (to get the number of digits of the power of two, if possible without the use of logarithms.



The idea came to me from this post



Using the fact that $5^3<2^7$ gives $10^168<2^560$.



Any advice to improve this bound and get the other are welcome.



Thanks






number-theory







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Using log it is between $10^168$ and $10^169$ though.
    $endgroup$
    – Hw Chu
    Mar 14 at 17:36










  • $begingroup$
    using wolfram $floorlog(2^561) +1= 169$
    $endgroup$
    – ahmed
    Mar 14 at 17:37







  • 2




    $begingroup$
    Which means that $10^169$ is the upper bound.
    $endgroup$
    – Hw Chu
    Mar 14 at 17:39










  • $begingroup$
    fixed thanks, the RHS inequality is apparently the trickier
    $endgroup$
    – ahmed
    Mar 14 at 17:41













1












1








1





$begingroup$


I would like to prove that $10^168 le 2^561< 10^169$ (to get the number of digits of the power of two, if possible without the use of logarithms.



The idea came to me from this post



Using the fact that $5^3<2^7$ gives $10^168<2^560$.



Any advice to improve this bound and get the other are welcome.



Thanks






number-theory







share|cite|improve this question











$endgroup$




I would like to prove that $10^168 le 2^561< 10^169$ (to get the number of digits of the power of two, if possible without the use of logarithms.



The idea came to me from this post



Using the fact that $5^3<2^7$ gives $10^168<2^560$.



Any advice to improve this bound and get the other are welcome.



Thanks






number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 14 at 17:40







ahmed

















asked Mar 14 at 17:25









ahmedahmed

324




324







  • 1




    $begingroup$
    Using log it is between $10^168$ and $10^169$ though.
    $endgroup$
    – Hw Chu
    Mar 14 at 17:36










  • $begingroup$
    using wolfram $floorlog(2^561) +1= 169$
    $endgroup$
    – ahmed
    Mar 14 at 17:37







  • 2




    $begingroup$
    Which means that $10^169$ is the upper bound.
    $endgroup$
    – Hw Chu
    Mar 14 at 17:39










  • $begingroup$
    fixed thanks, the RHS inequality is apparently the trickier
    $endgroup$
    – ahmed
    Mar 14 at 17:41












  • 1




    $begingroup$
    Using log it is between $10^168$ and $10^169$ though.
    $endgroup$
    – Hw Chu
    Mar 14 at 17:36










  • $begingroup$
    using wolfram $floorlog(2^561) +1= 169$
    $endgroup$
    – ahmed
    Mar 14 at 17:37







  • 2




    $begingroup$
    Which means that $10^169$ is the upper bound.
    $endgroup$
    – Hw Chu
    Mar 14 at 17:39










  • $begingroup$
    fixed thanks, the RHS inequality is apparently the trickier
    $endgroup$
    – ahmed
    Mar 14 at 17:41







1




1




$begingroup$
Using log it is between $10^168$ and $10^169$ though.
$endgroup$
– Hw Chu
Mar 14 at 17:36




$begingroup$
Using log it is between $10^168$ and $10^169$ though.
$endgroup$
– Hw Chu
Mar 14 at 17:36












$begingroup$
using wolfram $floorlog(2^561) +1= 169$
$endgroup$
– ahmed
Mar 14 at 17:37





$begingroup$
using wolfram $floorlog(2^561) +1= 169$
$endgroup$
– ahmed
Mar 14 at 17:37





2




2




$begingroup$
Which means that $10^169$ is the upper bound.
$endgroup$
– Hw Chu
Mar 14 at 17:39




$begingroup$
Which means that $10^169$ is the upper bound.
$endgroup$
– Hw Chu
Mar 14 at 17:39












$begingroup$
fixed thanks, the RHS inequality is apparently the trickier
$endgroup$
– ahmed
Mar 14 at 17:41




$begingroup$
fixed thanks, the RHS inequality is apparently the trickier
$endgroup$
– ahmed
Mar 14 at 17:41










1 Answer
1






active

oldest

votes


















3












$begingroup$

Using the inequality $(1+frac1x)^nx < e^n$ when $x > 0$, you get



$$
beginaligned
2^561 &= 2cdot (2^10)^56 = 2cdot10^168cdot (1+0.024)^56\
&< 2cdot 10^168 cdot e^1.344\
&< 2cdot 10^168 cdot 2.89^1.5 = 2cdot 10^168cdot 1.7^3\
&= 2cdot 4.913 cdot 10^168 < 10^169.
endaligned
$$






share|cite|improve this answer









$endgroup$












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    1 Answer
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    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Using the inequality $(1+frac1x)^nx < e^n$ when $x > 0$, you get



    $$
    beginaligned
    2^561 &= 2cdot (2^10)^56 = 2cdot10^168cdot (1+0.024)^56\
    &< 2cdot 10^168 cdot e^1.344\
    &< 2cdot 10^168 cdot 2.89^1.5 = 2cdot 10^168cdot 1.7^3\
    &= 2cdot 4.913 cdot 10^168 < 10^169.
    endaligned
    $$






    share|cite|improve this answer









    $endgroup$

















      3












      $begingroup$

      Using the inequality $(1+frac1x)^nx < e^n$ when $x > 0$, you get



      $$
      beginaligned
      2^561 &= 2cdot (2^10)^56 = 2cdot10^168cdot (1+0.024)^56\
      &< 2cdot 10^168 cdot e^1.344\
      &< 2cdot 10^168 cdot 2.89^1.5 = 2cdot 10^168cdot 1.7^3\
      &= 2cdot 4.913 cdot 10^168 < 10^169.
      endaligned
      $$






      share|cite|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        Using the inequality $(1+frac1x)^nx < e^n$ when $x > 0$, you get



        $$
        beginaligned
        2^561 &= 2cdot (2^10)^56 = 2cdot10^168cdot (1+0.024)^56\
        &< 2cdot 10^168 cdot e^1.344\
        &< 2cdot 10^168 cdot 2.89^1.5 = 2cdot 10^168cdot 1.7^3\
        &= 2cdot 4.913 cdot 10^168 < 10^169.
        endaligned
        $$






        share|cite|improve this answer









        $endgroup$



        Using the inequality $(1+frac1x)^nx < e^n$ when $x > 0$, you get



        $$
        beginaligned
        2^561 &= 2cdot (2^10)^56 = 2cdot10^168cdot (1+0.024)^56\
        &< 2cdot 10^168 cdot e^1.344\
        &< 2cdot 10^168 cdot 2.89^1.5 = 2cdot 10^168cdot 1.7^3\
        &= 2cdot 4.913 cdot 10^168 < 10^169.
        endaligned
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 14 at 17:49









        Hw ChuHw Chu

        3,302519




        3,302519



























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