2017年1月26日 · 1/e lnx=-1=>log_ (e)x=-1 =>e^ (-1)=x :.x=1/e

2015年7月4日 · You can use the definition of logarithm: logax = b → x = ab and the fact that ln = loge where e = 2.71828...: we can write: ln(ln(x)) = 1 ln(x) = e1 x = ee = 15.154

2015年11月14日 · Using the rules of logarithms, ln(x) + ln(x − 1) = ln(x ⋅ (x − 1)) = ln(x2 − x). Therefore, ln(x2 −x) = 1. Then, we exponentiate both sides (put both sides to the e power): eln(x2−x) = e1. Simplify, remembering that exponents undo logarithms: x2 −x = e. Now, we complete the square: x2 −x + 1 4 = e + 1 4 Simplify: (x − 1 2)2 = e + 1 4 = 4e +1 4 Take the square root of both ...

2016年7月11日 · =- 1/(x (ln x)^{2} ) you can do this simply as ( (ln x)^{-1})' =- (ln x)^{-2} (ln x)' =- (ln x)^{-2} 1/x =- 1/(x (ln x)^{2} ) if you want to fiddle about with e and ...

2015年2月23日 · For the first derivative start by rewriting 1 lnx = (lnx)−1 now take the derivative using power rule and chain rule dy dx = −(lnx)−2(1 x) = − 1 x(lnx)2 For the second derivative use the quotient rule. keep negative sign out in front so you do not lose track of it

2016年9月22日 · The graph of y is the standard function lnx shifted one unit down the y-axis. y has a zero at x=e f (x)=lnx has a vertical asymtote at x=0 and a zero at x=1 In this question y=f (x)-1 which simply shifts ("transforms") f (x) one unit down the y- axis.

2016年3月22日 · let y = elnx lny = lnelnx ->Take ln of both sides lny = lnx ⋅ lne -> use the property logbxn = nlogbx lny = lnx(1) -> lnee = 1 -> from the property logbb = 1 lny = lnx Therefore y = x

2018年3月18日 · 1) Divide each term by 2 ln(x) = 1 2 2) In solving x, we rewrite equation in logarithms form. eln(x) = e1 2 3) Exponential and base-e log are inverse function. x = e1 2 4) The value of x is e1 2 or 1.6487(4d.p.)

Using the chain rule: if y = ln(x) and u = 1 y then: du dx = du dy ⋅ dy dx So: d dx (1 lnx) = − 1 ln(x)2 ⋅ 1 x = − 1 xln(x)2

2017年5月30日 · This means the derivative of ln(lnx) is 1 x ⋅ lnx. This gives us the derivative of ln(lnx) ⋅ lnx which is lnx x ⋅ lnx + ln(lnx) x. If we do some cancellation we get: 1 x + ln(lnx) x, but since they both have denominators of x we can combine them to get ln(lnx) +1 x.