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

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 …

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 …

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 ...

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 …

Nov 14, 2015 · 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): …

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 …

May 23, 2018 · x=1/(e-1)~~0.582 Step 1 First, we must move all terms to one side. ln(1+x)-1-lnx=0 Step 2 We can now further simplify using the quotient rule. ln((1+x)/x)-1=0 Step 3 We …

Nov 30, 2015 · 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 ...