Recent content by guifb99

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    B A question regarding multiples of 3

    There, now I think it's WAY better understandable. I'm sorry. Lol
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    B A question regarding multiples of 3

    I'm sorry, I typed it wrongly, it's not "1/2", it's "1/24".
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    B Log(x), an easy and useful way to calculate it

    ½Logb(x2-1)≈logb(x) This is an easy and useful way to calculate the log of any natural number, including primes, it won't ever give a precise result, obviously (because of the -1), but as "x2-1" will always have divisors smaller than "x", you can easily calculate the approximation by using the...
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    B A question regarding multiples of 3

    Why doesn't any odd multiple of 3 can give "24-1(x2-1)=y" as a result (y) a natural number? Obviously no even number will make "y" a natural number, but all of the odd numbers do, but odd multiples of 3 (3, 9, 15, 21, 27...). x=1⇒y=0 x=3⇒y=1/3 x=5⇒y=1 x=7⇒y=2 x=9⇒y=10/3 x=11⇒y=5 x=13⇒y=7...
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    Easy and useful way to calculate Log(a+b)

    Yes, this is the second post I do about it, but now I did it in a better format, the other one was too confusing because I didn't know how to use the mathematical symbols in the thread, I'm new here.
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    Easy and useful way to calculate Log(a+b)

    a>b ⇒ Lim(a–b)→0 Logc(a+b) ≈ Logc(a) + Logc(√a⋅b) b>a ⇒ Lim(b–a)→0 Logc(a+b) ≈ Logc(a) + Logc(√a⋅b) a∈ℝ b∈ℝ c∈ℝ / c>0
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    I I found a *useful* method to calculate log(a+b), check it out

    Sorry, I'm new to this forum, didn't even know we could use these "uncommon outside the mathematic language" characters such as √ here, but thinking about it now, it makes sense that we would have these as this forum is basically about physics and mathematics. Anyway... thanks for explaining...
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    I I found a *useful* method to calculate log(a+b), check it out

    I don't even know what a geometric mean would be, I only know the basics of logarithms... I'm 15 years old
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    I I found a *useful* method to calculate log(a+b), check it out

    I'm brazilian so we use commas in cases like Log(2) = 0,3 But for americans, you can see the 0,3 as a 0.3 ;)
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    I I found a *useful* method to calculate log(a+b), check it out

    So, I found this method, I don't think I was the first to, though, but I don't see any post related to this anywhere on the internet, so maybe there's a slim chance I was the first? Anyway, it doesn't really matter. The method does not give the precise result, only approximations, but I find it...
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