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Nov 7, 2018 at 22:13. 5. "To log in" and "to log into" are Reflexive Separable Phrasal Verbs which often have the reflection omitted. They mean the same thing but have slightly different grammatical construction. "To log in" requires a prepositional phrase to describe what a person is logging into.
For my money, log on to a system or log in to a system are interchangeable, and depend on the metaphor you are using (see comment on your post). I suppose there is a small bit of connotation that "log on" implies use, and "log in" implies access or a specific user.
You could, however, do a change of base with the logs and put them in base 10. We have the formula logbx = logax logab where a can be any base you want. Most common base is 10. So we have, (log23)2 = (log103 log102)2 = (log3 log2)2 = log23 log22. Share.
the Taylor series for ln (x) is relatively simple : 1/x , -1/x^2, 1/x^3, -1/x^4, and so on iirc. log (x) = ln (x)/ln (10) via the change-of-base rule, thus the Taylor series for log (x) is just the Taylor series for ln (x) divided by ln (10). – correcthorsebatterystaple. Mar 18 at 14:35.
1. $\begingroup$. According to the international standard ISO 31-11 "ln" stands for base-e natural logarithm; "lg" is for base-10 common logarithm; and "lb" is for the base-2 binary one. "log" is a generic notation for a logarithm of an arbitrary base that needs to be specified.
2. There is currently no well-known function f(x, y) f (x, y) such that log(x) ⋅ log(y) = log(f(x, y)). log (x) ⋅ log (y) = log (f (x, y)). That is, the function f(x, y):= xlog(y) = ylog(x) f (x, y):= x log (y) = y log (x) has not been given a name yet, although it is a valid function. This situation may change at some future time.
Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
$\begingroup$ If so: no. Changing the base of the logarithm will make a difference by a constant factor; while $\log_2\log_2 x$ is exponentially smaller than $\log_2 x$. $\endgroup$ – Clement C. Commented Mar 6, 2017 at 19:24
Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.