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 $7^7^7^7$
$begingroup$
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
asked Mar 14 at 17:25
ahmedahmed
324
$endgroup$
add a comment |
$begingroup$
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
asked Mar 14 at 17:25
ahmedahmed
324
$endgroup$
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
add a comment |
$begingroup$
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
asked Mar 14 at 17:25
ahmedahmed
324
$endgroup$
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
asked Mar 14 at 17:25
ahmedahmed
324
asked Mar 14 at 17:25
ahmedahmed
324
edited Mar 14 at 17:40
ahmed
asked Mar 14 at 17:25
ahmedahmed
324
asked Mar 14 at 17:25
ahmedahmed
324
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Using the inequality $(1+frac1x)^nx < e^n$ when $x > 0$, you get
$$
beginaligned
2^561 &= 2cdot (2^10)^56 = 2cdot10^168cdot (1+0.024)^56\
&< 2cdot 10^168 cdot e^1.344\
&< 2cdot 10^168 cdot 2.89^1.5 = 2cdot 10^168cdot 1.7^3\
&= 2cdot 4.913 cdot 10^168 < 10^169.
endaligned
$$
answered Mar 14 at 17:49
Hw ChuHw Chu
3,302519
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Using the inequality $(1+frac1x)^nx < e^n$ when $x > 0$, you get
$$
beginaligned
2^561 &= 2cdot (2^10)^56 = 2cdot10^168cdot (1+0.024)^56\
&< 2cdot 10^168 cdot e^1.344\
&< 2cdot 10^168 cdot 2.89^1.5 = 2cdot 10^168cdot 1.7^3\
&= 2cdot 4.913 cdot 10^168 < 10^169.
endaligned
$$
answered Mar 14 at 17:49
Hw ChuHw Chu
3,302519
$endgroup$
add a comment |
$begingroup$
Using the inequality $(1+frac1x)^nx < e^n$ when $x > 0$, you get
$$
beginaligned
2^561 &= 2cdot (2^10)^56 = 2cdot10^168cdot (1+0.024)^56\
&< 2cdot 10^168 cdot e^1.344\
&< 2cdot 10^168 cdot 2.89^1.5 = 2cdot 10^168cdot 1.7^3\
&= 2cdot 4.913 cdot 10^168 < 10^169.
endaligned
$$
answered Mar 14 at 17:49
Hw ChuHw Chu
3,302519
$endgroup$
add a comment |
$begingroup$
Using the inequality $(1+frac1x)^nx < e^n$ when $x > 0$, you get
$$
beginaligned
2^561 &= 2cdot (2^10)^56 = 2cdot10^168cdot (1+0.024)^56\
&< 2cdot 10^168 cdot e^1.344\
&< 2cdot 10^168 cdot 2.89^1.5 = 2cdot 10^168cdot 1.7^3\
&= 2cdot 4.913 cdot 10^168 < 10^169.
endaligned
$$
answered Mar 14 at 17:49
Hw ChuHw Chu
3,302519
$endgroup$
Using the inequality $(1+frac1x)^nx < e^n$ when $x > 0$, you get
$$
beginaligned
2^561 &= 2cdot (2^10)^56 = 2cdot10^168cdot (1+0.024)^56\
&< 2cdot 10^168 cdot e^1.344\
&< 2cdot 10^168 cdot 2.89^1.5 = 2cdot 10^168cdot 1.7^3\
&= 2cdot 4.913 cdot 10^168 < 10^169.
endaligned
$$
answered Mar 14 at 17:49
Hw ChuHw Chu
3,302519
answered Mar 14 at 17:49
Hw ChuHw Chu
3,302519
answered Mar 14 at 17:49
Hw ChuHw Chu
3,302519
answered Mar 14 at 17:49
Hw ChuHw Chu
3,302519
3,302519
add a comment |
add a comment |
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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