2017年5月30日 · So, we turn #(lnx)^(lnx)# into #e^((ln(lnx))(lnx))#. This works because, due to raising e to any power is an inverse function to taking the natural log of any number, #e^ln(x)=x# for any x. This may seem like a more complicated function we have made to take the derivative, but trust me, it makes thins simpler.

2016年3月22日 · e^lnx=x let y=e^lnx ln y=lne^lnx->Take ln of both sides lny = lnx * ln e -> use the property log_b x^n ...

2017年1月26日 · 1/e lnx=-1=>log_(e)x=-1 =>e^(-1)=x :.x=1/e. 141889 views around the world ...

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

2018年4月3日 · Both! ln(x) + ln(x) = ln(x^2) = 2ln(x) Sum of logarithms with same base is equivalent to the logarithm of the product. So: ln(x) + ln(x) = ln(x^2) It also turns out that: ln(x^2) = 2ln(x) So technically: ln(x) + ln(x) = ln(x^2) = 2ln(x) You can also just treat the ln(x) like it's a variable and sum them. If you have ln(x) + ln(x), then you have 2ln(x).

2018年3月20日 · Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e. 1 Answer . Rhys

2018年5月19日 · #y=lnx^(lnx)=e^(ln(lnx^(lnx))# #(y)'=(e^(ln(lnx^(lnx))))'# #=# #e^(ln(lnx^(lnx)))*(ln(lnx^(lnx)))'# #=# #lnx^(lnx)*(lnx(ln(lnx))'# #=#

x = 1/e^5 From the definition of a logarithm, we have the property e^ln(x) = x. From this: ln(x) = -5 => e^(ln(x)) = e^(-5) => x = 1/e^5

2017年6月30日 · See below. f(x) = (lnx)^2 lnx is defined for x>0 hence, f(x) is defined x>0 lim_(x-> 0) f(x) = +oo and lim_(x->oo) f(x) =+oo f'(x) = 2lnx*(1/x) {Chain rule] For a ...

2016年2月6日 · #color(brown)("Total rewrite as changed my mind about pressentation.")# #color(blue)("Preamble:")# Consider the generic case of #" "log_10(a)=b#