Mar 2, 2018 · -1 Division rule of logarithms states that: ln(x/y) = ln(x) - ln(y) Here we can substitute: ln(1/e)=ln(1) - ln(e) 1) Anything to the power 0=1 2) ln(e)=1, as the base of natural logarithms is always e Here, we can simplify: ln(1)=0 ln(e)=1 Thus: ln(1)-ln(e)=0-1 =-1 Thus, we have our answer

Jun 10, 2016 · It is -1. We apply the properties of the logarithm: ln(1/e)=ln(e^(-1)) the first property is that the exponent "exit" and multiply the log ln(e^-1)=-ln(e) the second property is that the logarithm of the base is 1. The base of the natural logarithm is e then …

Mar 23, 2016 · This expression is not valid because #ln# function is only defined for positive numbers. Explanation: This expression is not valid because #ln# is not defined for negative numbers but if it was:

Dec 3, 2015 · ln(1/e^(3))=ln(e^(-3))=-3. Since ln(x) and e^{x} are inverse functions, ln(e^{x})=x for all values of x. Since 1/e^{3}=e^{-3} by definition of negative exponents, it ...

Question 657977: find the exact value of the expression ln(1/e) Answer by jim_thompson5910(35256) ( Show Source ): You can put this solution on YOUR website!

Feb 25, 2017 · From here, use #ln(a/b)=ln(a)-ln(b)# and #ln(e^x)=x#: #=ln(e^x-1)-ln(e^x)=color(blue)(ln(e^x-1)-x# I don't know if this is a simplification per se, but it's definitely a valid way to rewrite the function.

May 31, 2018 · How do you simplify #Ln(1/e^2)#? Precalculus Properties of Logarithmic Functions Natural Logs. 1 Answer

Jan 1, 2016 · Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e. 1 Answer . Guilherme N.

Apr 26, 2015 · The code to render this properly is f(x)=ln((1+e^x)/(1-e^x)) (notice the usage of parentheses). This is a function composition. Given two functions #y=g(x)# and #z=h(y)# you can compute the derivative of the composition #f(x)=h(g(x))# as follows:

Apr 24, 2016 · f'(x)=e^x/(1+e^x) The chain rule says that if f(x)=g(h(x)) then f'(x)=h'(x)g'(h) Substituting in the functions from the question, h(x)=1+e^x g(h)=ln(h) Then the derivatives are h'(x)=e^x g'(h)=1/h According to the chain rule, f'(x)=(e^x)*(1/(h(x))) Expanding out and substituting in that h(x)=1+e^x, f'(x)=e^x/(1+e^x)