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 a 5 5...how do you solve this? Thank you.

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 intended to teach in class is that the statement "For all x, y x, y, ln(x + y) = ln(x) + ln(y) ln (x + y) = ln (x) + ln (y) " is false. Equivalent, that the statement "There exist x, y x, y such that ln(x ...

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 / log 2 is irrational. Suppose on contrary that log 3/ log 2 = p/q log 3 / log 2 = p / q where p and q are integers. Since 0 <log 3/ log 2 0 <log 3 / log 2 , we can choose p …

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 of 20 it has a value...

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 of natural numbers. As increasing any of the terms in the original expression can only cause the resulting value to be greater, this new expression can be used to calculate an upper bound for the ...

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 it is (relatively) easy to find out how to differentiate eu, and we can then use the knowledge of that derivative to figure out the derivatives of the exponential functions of other bases. Intuitively, the reason you should ...

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.