Solve x(lnx)=1.

You can consider in this way:


x ln x = 1

ln ( x^x ) = 1

ln ( x^x ) = ln e

x^x = e [ not x = e here, remember ]


Of course it has an unique solution for it is a continuous functions and we must have a root between 1.6 and 2, for 1.^1.6 < e and 2^2 = 4 > e.


However, there is no answer which can be expressed in Surd Form. What you can do is to work out a very close value by Bisection Method, a method which has been removed from HKCEE syllabus since 2006. You can use it to find an approximate value correct to any decimal places you like.


Hope I can help you.



2011-02-27 17:05:45 補充：
For sure Yuk should understand now how the answer can be obtained and why is the original tackling approach unable to find the solution.

Solve x(lnx)=1.

lnx = 1/x

If you familiarize with the graph of ln x and 1/x, then it is not difficult to see that there is only one solution.

On the other hand, consider f(x) = x(lnx) - 1

f(1) = -1 and f(2) = 0.3862

By intermediate value theorem, there should be at least one root in the interval (0.5,2)

Also, f'(x) = (lnx) + 1 which is strictly increasing for x > 1

Since f(1) = -1 and f(x) -> infinity when x -> infinity and f(x) is a continuous function. There should be at least one root by intermediate value theorem

To find out the solution, you can use the bisection method or Newton's method.

1.763222834

Consider Newton's method