Videos
How do I calculate the logarithm in base 2?
To calculate the logarithm in base 2, you probably need a calculator. However, if you know the result of the natural logarithm or the base 10 logarithm of the same argument, you can follow these easy steps to find the result. For a number x:
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Find the result of either
log10(x)orln(x). -
Divide the result of the previous step by the corresponding value between:
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log10(2) = 0.30103; or -
ln(2) = 0.693147.
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The result of the division is
log2(x).
What is the logarithm in base 2 of 256?
The logarithm in base 2 of 256 is 8. To find this result, consider the following formula:
2x = 256
The logarithm corresponds to the following equation:
log2(256) = x
In this case, we can check the powers of 2 to see if we can find the value of x: 20 = 1, 21 = 2, 22 = 4, …, 27 = 128, and 28 = 256.
Since we found the argument of our logarithm, we can write that:
log2(256) = 8.
Why is the logarithm in base 2 important?
In a computer world, binary code is of essential importance: words, numbers, pictures, and everything else can be reduced to a string of 0s and 1s. Since the binary code uses only two digits, the number 2 appears consistently in computer science.
The widespread appearance of log2 in computer science has no strong mathematical reason (since logarithms can change base by multiplication) but can be useful. For example, using log2 to compute entropy allows us to obtain the result expressed in bits, which are the natural unit.
I got this question from this video and there were 4 options:
https://www.youtube.com/watch?v=X6C5hGpWW5A
can anyone tell me if my method was accurate, (the guy in the video used a more simpler method but here's mine):
we know 2>x>1 (2^2 = 4 >3 > 2^1 =2) then I took three values as x from the question:
(i) 2^ 3/2 = 2 *sq rt (2)
(ii) 2 ^ 4/3 = 2* cub rt (2)
(iii) 2 ^ 5/3 = 2* cub rt (4)
As (ii) n (iii) are the closest values to 3, log base 2 of 3 must lie between 1 1/2 and 1 2/3. How accurate do you think this method would be? I will really really enjoy some feedback it has really been bugging me.