Apr 5, 2015 · The answer is 0. ln(1) is the same as asking e to what power is 1? Since anything to the 0 power is 1, ln(1) = 0

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

Aug 30, 2015 · Although you could use #d/dx (ln(u)) = 1/u (du)/dx#, the algebra will get messy that way. Let's rewrite using properties of #ln# . #y = ln(1+(1/x)) = ln((x+1)/x)#

Aug 20, 2016 · (x+1)ln(1+x)-x+C We have: I=intln(1+x)dx We will use integration by parts, which takes the form: intudv=uv-intvdu So, for intln(1+x)dx, let: {(u=ln(1+x)" "=>" "du=1 ...

Feb 6, 2017 · ln(1+x) = sum_(n=0)^oo (-1)^n x^(n+1)/(n+1) with radius of convergence R=1 Start from: ln(1+x) = int_0^x (dt)/(1+t) Now the integrand function is the sum of a geometric series of …

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 …

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 …

Apr 26, 2018 · Then, dv=1 \ dx, v=int1 \ dx=x. We don't put the constant until we finish the whole integration. Inputting, we get, intln(1/x) \ dx=xln(1/x)-intx*(-1/x) \ dx =xln(1/x)-int-1 \ dx …

Dec 8, 2015 · Hence, even though the radius of convergence is 1, the series for ln(1-x) converges and equals ln(1-x) over the half-open/half-closed interval [-1,1) (it doesn't converge at x=1 …

May 7, 2016 · How do you find the Limit of #ln(n+1) - ln(n) # as n approaches infinity? Calculus Limits Determining Limits Algebraically. 1 Answer