How to solve equations of this form: $x^x = n$? How would I go about solving equations of this form:
$$
x^x = n
$$
for values of n that do not have obvious solutions through factoring, such as $27$ ($3^3$) or $256$ ($4^4$).
For instance, how would I solve for x in this equation:
$$x^x = 7$$
I am a high school student, and I haven't exactly ventured into "higher mathematics." My first thought to approaching this equation was to convert it into a logarithmic form and go from there, but this didn't yield anything useful in the end.
My apologies if this question has been asked and answered already; I haven't been able to find a concrete answer on the matter.

you can solve it using newton's method  
$$ f(x) = x^x - 7 = 0 \Rightarrow f'(x)  = x^x(\ln x+ 1) $$
now choose $  x_0  $ and let it be $ x_0 = 2 $
and use the formula $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}  = x_n - \frac{x_n^{x_n} - 7}{x_n^{x_n}(\ln x_n + 1) }$$
now just evaluate $ x_1 $ by using $ x_0 $ then $ x_2 $ then $x_3 \cdots $  by a calculator and you'll find an approximation 
$$ x_1 \approx 2.442962082 $$
$$ x_2 \approx 2.331852211 $$
$$ x_3 \approx 2.316698614 $$
$$ x_4 \approx 2.31645502 $$
$$ x_5 \approx 2.316454959 $$
$$ x_6 \approx x_5 $$
$$ x_7 \approx x_6 $$
$$ x \approx 2.316454959 $$
http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/approx/newton.html
http://mathworld.wolfram.com/NewtonsMethod.html
