I would like to know how logarithms are calculated by computers. The GNU C library, for example, uses a call to the fyl2x() assembler instruction, which means that logarithms are calculated directl...

2020年8月14日 · For example, to calculate $\log _2 3$ with $9$-digit precision, you must iterate at least $30$ times. This is NOT a restriction due to the precision of floating-value calculation; even if you calculate by hand or with arbitrary precision float (i.e. so …

2017年4月23日 · How to calculate non-exact Log problems like: Base 2 Log 10 = 3.32192809489? See more linked questions ...

If your number is close to 1, then the natural log will be a small number, and multiplication by the natural log of 10 might be less of a headache. Of course, this does not cover the range of numbers you want; you must use RANGE REDUCTION. To do this, first calculate, by any means available, the base 10 logs of 1/0.5, 1/0.9, 1/0.99, and 1/0.999.

***** value to take the log factorial of is 1 Exact value is 0.0 Small value approximation (K.V. Raman) is -1.0 Stirling is -0.0810614667953 Stirling with extra term is -0.00101875912179 Ramanujan is -0.000143497498377 ***** ***** value to take the log factorial of is 2 Exact value is 0.69314718056 Small value approximation (K.V. Raman) is 0. ...

2016年2月14日 · Hence, to calculate $\ln n$ in practical applications, first calculate $\log_{20} n$, then multiply it by $3$. Since $20$ is an integer, it's easier to work with. For example, if we need to calculate $\ln 34 627 486 221$, we can do the following:

2015年6月7日 · $$\log_{10}(53) = 2 -1 + \frac{53}{100-10}=1.588888$$ Logarithm maps numbers from 10 to 100 in the range 1 to 2 so log of numbers near 50 is about 1.5. But this is only a linear approximation, good for mental calculation and toy projects but not that good for serious research.

I had rather only do first order approximations on a better decomposition of $4$ (also using powers of $5$ and even $7$ for example) or calculate many terms only using $\operatorname{ln}(4)=4\operatorname{argth}\left(\frac13\right)$, getting one decimal per each new easily calculated term in the series (though you need to compute a division for ...

2018年3月18日 · You can then compute $$ \log_2(256!)=\frac{\log\Gamma(257)}{\log 2} $$ with minimal difficulty (and without leaving double precision). In general some care is needed to work with the gamma function (branch cuts and numerical analysis), but in your case as long as you stick to factorials of numbers from 1 to 10 305 or so you should be just fine ...

If you memorize more logarithms of small integers and/or memorize $\log_2 5$ to more precision you can do even better. For example, suppose you want to estimate $\log_2 75$. This is $\log_2 3 + \log_2 25 \approx \log_2 3 + 4.6$, so you're reduced to estimating $\log_2 3$. Well, $$\log_2 3 = \log_2 15 - \log_2 5 \approx \log_2 16 - 2.3 = 1.7$$