Express ln 32−−√3 ln 32 3 in terms of ln 2 ln 2 and/or ln 3 ln 3 My try: ln 32−−√3 = ln(321/3) = 1 3ln 32 = 1 3ln25 ln 32 3 = ln (32 1 / 3) = 1 3 ln 32 = 1 3 ln 2 5 This isn't correct because there's …

2021年11月26日 · Your example shows that ln(x + y + z) ≠ ln(x) + ln(y) + ln(z) ln (x + y + z) ≠ ln (x) + ln (y) + ln (z) (assumed to be stated for all values of x, y, z x, y, z) is false. What they really …

I think I have to use the arcsin(x) arcsin (x) formula. Which means would I use y =u2 y = u 2 with u = e−x u = e − x but that does not seem to work.

Solve the following If f(x) =e−x + 2e−2x + 3e−3x + ⋯ f (x) = e − x + 2 e − 2 x + 3 e − 3 x + ⋯ Then find ∫ln 3 ln 2 f(x)dx ∫ ln 2 ln 3 f (x) d x I don't have any idea.

2017年8月22日 · The target value is ln3 = 1.098612289…, with which we agree to the precision displayed after only four terms in the series and only computing x to six places.

2023年10月14日 · I was trying to figure out if ln(3)/ ln(2) ln (3) / ln (2) is transcendental, when I found this post by b_jonas But there's a proof just as simple showing that log 3/ log 2 log 3 / …

2018年5月9日 · I was playing around with nested radicals and I decided to see if nested equations of logarithms would converge. It seems to converge to a value around $1.368$, and at a depth …

Since this is a super-exponential series and the gap between the first two terms is greater than 1 1 it is safe to conclude that every term in the series is greater than the same term in the series …

2015年10月26日 · The reason it's so much more useful to write 3x =ex ln 3 rather than 3x =3xlog3(3) or 3x = 10xlog10(3) or any other variation using logarithms of different bases is that …

2019年3月4日 · Our teacher tells us to convert it this way 3x =eln3x =ex⋅ln 3 and then use the rule eu ⋅ u′ but I can't understand where ln comes from and how ln3x = x ⋅ ln 3.