Number of roots for x^x - m=0 respectively to the change of values of m.?

Question

i have already solved for x>0 the answer is e^(-1/e). can someone helps with me x<0. dont know how to do. the positive i did with derivative and it worked. and i know the answer is right because the book has given the answer to help us but of course we have to find the method. the answer that it has given is 1.....

thanks in advance.

Answer 1

I don&#039;t understand what you are asking. The number of roots clearly depends on m and the field you are solving over.

For m = 0, there are clearly no roots.

For m &gt; 0, there is only one real root.

For x &gt; 0 the equation is equivalent to x ln x = ln m

Over the reals, ln x is monotonic and so is x, so x ln x is monotonic too. This means there is only one root. So what you mean by e^(-1/e) is not at all clear.

Over the complex plane, let x = r e^(i@)

then ln(x^x) = x ln(x) = (r e^(i@)) (ln(r) + i@)

so ln(x^x) = r (cos @ + i sin @)(ln(r) + i@) = r ((cos @ - @ sin@) + i(ln(r) + @ cos @)

So for any given r, there is a value of @ such that:

ln(r) + @ cos @ = 0

so for that x = r e^(i@), m = r (cos @ - @ sin@) will have x ln(x) = ln(m) which implies that there are two roots for (at least portions of) the complex plane.