SOLUTION: I need help solving the following... ln(x)^(ln(x))=4 so far I took both sides times "e" to get... e^ln(x)^(ln(x))=e^4 x^(ln(x))=e^4 but I don not know what to do next!! T

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I need help solving the following... ln(x)^(ln(x))=4 so far I took both sides times "e" to get... e^ln(x)^(ln(x))=e^4 x^(ln(x))=e^4 but I don not know what to do next!! T      Log On


   

Question 5155: I need help solving the following...
ln(x)^(ln(x))=4
so far I took both sides times "e" to get...
e^ln(x)^(ln(x))=e^4
x^(ln(x))=e^4
but I don not know what to do next!!
Thanks for the help
Found 2 solutions by khwang, rapaljer:
Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
Given ln(x)^(ln(x))=4
Let y = ln x, the equation becomes
y^y = 4,
But y^y = 2^2, [This is the point: to take advantage 4 = 2^2]
so y = 2 = ln x,
We get x = e^2
It seems a little tricky. While there is no other good
ways even though I have tried to take ln , log2 , etc.

Your idea is good, but not working for solving this tricky problem.
Kenny

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Let me take a shot at this, since I found TWO solutions for it.
ln+%28x%5E%28ln+x%29%29+=+4

Using the second law of logarithms, i.e. log+%28M%5EN%29+=+N+%2A+log+M+, this can be written
ln+%28x%5E%28ln+x%29%29+=+4
%28ln+x%29+%2A+%28ln+x%29+=+4 or
%28ln+x%29%5E2+=+4+

Take the square root of each side of the equation:
+ln+x+=++2+ or ln+x+=+-2

In each case above, raise both sides as a power of e:
e%5E%28ln+x%29+=+e%5E2 or e%5E%28ln+x%29+=+e%5E%28-2%29+
+x+=+e%5E2 or x+=+e%5E%28-2%29+

The only question that remains is are these solutions both acceptable. As far as I can tell, the value of x=e%5E%28-2%29+ never makes the ln of a negative, so BOTH answers should be included in the solution.

R^2 at SCC