Question

the number of real roots of xlnx-1= zero is​

Answer 1

Answer:

x>1( read the solution)

Step-by-step explanation:

[Only real solutions]

Cuz ln(x) should be valid, x should be positive. right?

Then, (x-ln(x))’ = 1 - 1/x > 0 for x > 1. Thus, x-ln(x) is strictly increasing for x >= 1(you can prove it by Mean value theorem). Then, x-ln(x) = 1 has at most one solution for x >= 1.(Why? strictly increasing function increases as x gets increased. Thus, if x - ln(x) = 1 has two different solutions on [1, Inf), that’s weird. Let those x1 and x2. Then, w.l.o.g. 1=<x1 < x2. Then, x1-ln(x1) < x2-ln(x2)! Contradiction!) But, x+ln(x) = 1 holds when x = 1. Thus, x = 1 is the unique solution for x>=1.

Then, what if x < 1? Then, (x-ln(x))’ = 1 - 1/x < 0. Thus, x - ln(x) is strictly decreasing for x<=1. Then, you can use the same logic above and you can conclude that x =1 is the unique real solution for the equation.

[Complex solutions]