2014年12月15日 · Lets start by breaking down the function. (ln(x))/x = 1/x ln(x) So we have the two functions; f(x) = 1/x g(x) = ln(x) But the derivative of ln(x) is 1/x, so f(x) = g'(x). This means we can use substitution to solve the original equation. Let u = ln(x). (du)/(dx) = 1/x du = 1/x dx Now we can make some substitutions to the original integral. …

2007年12月8日 · How do you go about deriving the series expansion of ln(x)? 0 < x I got the representation at math.com but i'd still like to know how they got it. It's been a while since i did calc. iii. Thanks. John

2015年11月14日 · Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e. 1 Answer . mason m

2018年3月17日 · See below. Using the exponential at both sides as the inverse of ln we obtain e^x = x but y = e^x and y = x does not intersect so no real solution for x = 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

I found: x=e^e=15.154 You can use the definition of logarithm: log_ax=b->x=a^b and the fact that ln=log_e where e=2.71828...: we can write: ln(ln(x))=1 ln(x)=e^1 x=e^e=15.154

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

2015年5月20日 · firstly we look at the formula for the Taylor series, which is: f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n which equals: f(a) + f'(a)(x-a) + (f''(a)(x-a)^2)/(2!) + (f ...

2015年11月11日 · It is exactly x. You are looking for a number that is the exponent of the base of ln which gives us the integrand, e^x; so: the base of ln is e; the number you need to be the exponent of this base to get e^x is.....exactly x!!! so: ln(e^x)=log_e(e^x)=x

2015年10月23日 · It's x. The logarithm and the exponential are inverse function, which means that if you combine them, you obtain the identity function, i.e. the function I such that I(x)=x. In terms of definitions, it becomes obvious. The logarithm ln(x) is a function which tells you what exponent you must give to e to obtain x. So, e^(log(x)), literally means: "e to …