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I don't know if the landlord missed the conditions in the question (just guess) There is a condition for the general exam, that is, LG2 = there will be a logarithm of 10 to get 100LG2 = so 2 100 = 10
That is, 2 100 is to the power of 10.
10 to the 30th power is 31 digits, 10 to the 31st power is 32 digits 100....0 So 2 100 is at least 31 digits, and more than 32 digits of 100....0 small.
So it's 31 digits.
This is the case in formal exams, whether it's a few squares, or how many powers.
Above Xiao Yao Ruyi got the right answer, but there was no right process.
He only stated that it was at least 31 digits, but he really couldn't say that 1024 was not carried with 10 consecutive digits.
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2^100=(2^10)^10=(1000+24)^10=c<10,0>*1000^10*24^0+c<10,1>*1000^9*24^1+
c<10,2>*1000^8*24^2+..c<10,10>*1000^0*24^10
There are 11 items in total, of which c is capitalized c, the subscript is a, the superscript is b, the first item is 31 digits, and the first item is 1
The second item starts with each item with less than 30 digits
The sum of the 11 entries will not be rounded up to 32.
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1024 1024, take a vertical look, the product is 4 + 3 = 7 digits.
Then every time you multiply by 1024, the number of digits of the product increases by 3 digits.
1024 10, the number of digits is: 3 (10-1)+4=31 digits.
So 2 100 is 31 digits.
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log(2)=
x=2^100
log(x)=100*log(2)=
x=10 x=10 1=10 ==>(1+1) = 2 digits.
x=10 31=1*10 31 ==>1 after 31 0 ==>(31+1)=32 digits.
x=10 >(30+1)=31 digits.
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2^10=10242^100=(2^10)^10=(1000+24)^10=c*1000^10*24^0+c*1000^9*24^1+c*1000^8*24^2+.+c*1000 0*24 10 a total of 11 items, of which c is uppercase c, subscript is a, superscript is blind daunting b, the first item is 31, and the first place is 1, the second grinding item starts with the number of digits of each item is small and late at 3011 items and will not enter.
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2 times out of 10 is 1024
100 times for 2 is 10 times for 1024.
1024 is about 1*10 3
Therefore, the 100 times of 2 is about a single digit attack on the oak number *10 30, which is just like a sign is 31 bits.
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1. The standard year-old closed computer calculates:
2 100 = 1267650600228229401496703205376 it has 31 posture numbers.
2. No computer, manual algorithm is as follows.
1024 10>1000 10=10 30 This is a 31-bit trace of the data.
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Let x=2 100, then logx=log(2 100)=100*log2=so x=10
Because 10 30 and because 10 30 is a 31-digit number, and 10 31 is a 32-digit number, x is a 31-digit number.
So 2 100 is 31 digits.
If you don't want to do the math, you'll probably have to.
Note: 2 100 means 2 to the power of 100.
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I learned logarithms in high school. 100LG2 = means that 2 to the power of 100 is 31 digits.
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2^100=(2^10)^10
We know that 2 10 = 1024 1
1048576 5 (When I calculated this step, I wanted to look at 1024*1024 when it was a few digits, and I wanted to see that 48576 can be approximated to 1000000 compared with 1048566, and it can be rounded.) This is of great help to the following research. )
The following calculation is the multiplication of 5 seven-digit numbers.
In the same way, 7 digits multiplied by 7 digits is 13 digits (followed by 3 7 digits), 13 digits multiplied by 13 digits is 25 digits, and 25 digits multiplied by 7 digits are 31 digits.
So. The number of 2,100 is 31.
I didn't pass the computer verification, so that's how it should be. )
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Because 2 10 = 1024
The number of digits is the same as the number of 1000*1000 digits, which is 6 digits.
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Can it be solved with logarithmic knowledge?
lg2=, so lg2 100=100lg2=, i.e. 2 100=10, so it's a 31-digit number.
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1. The standard computer calculates:
2 100 = 1267650600228229401496703205376 it has 31 digits.
2. No computer, manual algorithm is as follows.
1024 10>1000 10=10 30 This is a 31-bit data.
i.e. 10 30<2 100<371293*10 25
Since 10 30 and 371293*10 25 are both 31-bit data, then 2 100 is also a 31-bit data.
Note: 4 5 vs. 13 5 This can be calculated by hand, and it doesn't take long to calculate.
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Let x=2 100, then logx=log(2 100)=100*log2=Mingju....The big locust bureau rolls away.
So x=10
Because 10 30
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