2017年9月23日 · Yes, but also see below ln^2 x is simply another way of writing (lnx)^2 and so they are equivalent. However, these should not be confused with ln x^2 which is equal to 2lnx There is only one condition where ln^2 x = ln x^2 set out below. ln^2 x = ln x^2 -> (lnx)^2 = 2lnx :. lnx * lnx = 2lnx Since lnx !=0 lnx * cancel lnx = 2 * cancel lnx lnx = 2 x =e^2 Hence, ln^2 x = ln x^2 is only true for x=e^2

2016年4月3日 · (ln(2x))' = 1/(2x) * 2 = 1/x. You use the chain rule : (f @ g)'(x) = (f(g(x)))' = f'(g(x)) * g'(x). In your case : (f @ g)(x) = ln(2x), f(x) = ln(x) and g(x) = 2x.

2016年8月25日 · ln e^(2x) = 2x As a Real valued function, x |-> e^x is one to one from (-oo, oo) onto (0, oo). As a result, for any y in (0, oo) there is a unique Real value ln y such that e^(ln y) = y. This is the definition of the Real natural logarithm. If t in (-oo, oo) then y = e^t in (0, oo) and from the above definition: e^(ln(e^t)) = e^t Since x |-> e^x …

2014年7月24日 · This is the composite of ln x and 2x, so we use the Chain Rule together with the facts that (2x)'=2 and that (ln x)'=1/x: (ln(2x))'=1/(2x) \times (2x)'=2/(2x)=1/x.

2015年11月30日 · x = 2/e We will use the following: ln(a^x) = xln(a) ln(a) - ln(b) = ln(a/b) e^ln(a) = a 2ln(x) + 1 = ln(2x) => ln(x^2) + 1 = ln(2x) (by the first property above ...

2015年5月25日 · What is the derivative of #f(x)=sqrt(1+ln(x)# ? What is the derivative of #f(x)=(ln(x))^2# ? See all questions in Differentiating Logarithmic Functions with Base e

2016年11月24日 · x(ln(2x)-1)+C >I=intln(2x)dx We should use integration by parts in the absence of all other possible integration strategies.

2014年12月14日 · Integration by Parts int u dv=uv - int v du int ln(2x+1) dx by Substitution t=2x+1. => {dt}/{dx}=2 => dx={dt}/2 =1/2int ln t dt by Integration by Parts, Let u=ln t and dv=dt => du = 1/{t}dt" " v=t =1/2(t ln t - int dt) =1/2(t ln t - t) + C by putting t=2x+1 back in, =1/2[(2x+1)ln(2x+1)-(2x+1)]+C I hope that this was helpful.

2016年8月6日 · x = sqrt(2)/2e If x > 0 then: ln x + ln(2x) = ln (x*2x) = ln(2x^2) and 2 = ln e^2 So we have: ln (2x^2) = ln (e^2) Since ln is one-one as a Real valued function of positive Reals, this implies that: 2x^2 = e^2 So: x^2 = e^2/2 Hence: x = e/sqrt(2) = (sqrt(2))/2e Note we ignore the negative square root since we are only looking at the case x > 0.

2015年6月2日 · int ln(2x+7)dx= int 1* ln(2x+7)dx. Now integrate by parts, ln(2x+7) * x - int 2/(2x+7) *x dx = x ln(2x+7)- int (2x)/(2x+7) dx = x ln (2x+7) - int (1- 7/(2x+7) )dx = x ...