# 1 ln x

See other nonelementaryhttps://www.youtube.com/watch?v=0aN4lSSWKkANote: I forgot to write the 5th term in terms of x.Thanks Gabriel!Gabriel Shapiro: You forg

Now, we follow the same process to get rid of this ln to get: e^(ln(x)) = e^e. And therefore: x = e^e <=== FINAL ANSWER. I hope that helps you out! Please let me know if you have any other questions! I = 1 2 ln(x 2 + 1) ln(x).

Near x=0, I suppose you could expand e[sup]x[/sup], in terms of a series expansion, and then discard higher order terms, and then expand the resulting log, again in terms of a power series. (x-1)/(x^2-1)=0 returns the message no solution, domain definition is taken into account for the calculation, the numerator admits x = 1 as the root but the denominator is zero for x = 1 , 1 can't be a equation solution. The equation does not admit a solution. equation_solver(1/(x+1)=3) returns [-2/3] Solving quadratic equation online 1 y + 1 x = ln 2 y − 1 1 x = (ln(y + 1) − ln(y − 1)) 2 There are many equivalent correct answers to this question.