- verbs - log in to or log into or login to - English Language . . .
As for “Log in to host com” versus “Log into host com,” I would use the former because I think that “log in” is a fixed phrase Martha’s answer to another question is also related Added : The Corpus of Contemporary American English (COCA) lists 65 occurrences of “log in to” and 58 occurrences of “log into,” both including
- What is the difference between log in, sign in; register, sign up; log . . .
log in, sign in, login, log on, logon, authenticate All of these words are more-or-less synonyms "Login" "logon" may or may not be acceptable (check your style guide) If they are, they may be nouns instead of verbs (referring to the action of logging in or the state of being logged in)
- When log is written without a base, is the equation normally referring . . .
$\log (x)$ refers to $\log_2 (x)$ in computer science and information theory $\log(x)$ refers to $\log_e(x)$ or the natural logarithm in mathematical analysis, physics, chemistry, statistics, economics, and some engineering fields $\log(x)$ refers to $\log_{10}(x)$ in various engineering fields, logarithm tables, and handheld calculators
- Logged-in, log-ined, login-ed, logined, log-in-ed, logged in?
@Paul: As the past tense of the verb, it would be logged in, as in I logged in this morning As an adjectival phrase, it could be either logged in or logged-in, typically depending on placement, e g
- Easy way to remember Taylor Series for log(1+x)?
$\begingroup$ I think something is wrong with the derivation you have - notably, the first equation, $\log(1-x)=-\sum_{n=1}^{\infty}x^n$ is not true - you probably want a log around the sum on the left $\endgroup$
- The difference between log and ln - Mathematics Stack Exchange
$\begingroup$ Since the default base of log can vary between and even within fields, seems a good rule of thumb is to treat ln as loge (of course), and log as unknown (re: base-2 10 e whatever) until you confirm the context If calculating or programming, check a test result before making assumptions
- Taylor Series for $\\log(x)$ - Mathematics Stack Exchange
$\begingroup$ @Kurtoid No because when you plug in $1-x$ into the alternating sign expansions, for odd exponents a negative sign is introduced, for evens nothing changes, so $\log(1-x)$ is simply just $-x - \frac{x^2}2 - \frac{x^3}3 - \dots$ $\endgroup$
- Intuition behind logarithm inequality: $1 - \\frac1x \\leq \\log x . . .
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