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.

Can be solved by solving the problem of finding an approximate value.But accuracy ...
$(x_0+\triangle x)^{x_0+\triangle x}=7\\x=2;\triangle x=?\\f(x)=f(x_0)+f'(x_0)\triangle x\\f(x_0)=2^2=4\\f'(x)=x^x(lnx+1)*\triangle x\\7=4+4(ln2+1)*\triangle x\\\triangle x=\frac{3}{4(ln2+1)}\approx0.44\\x=2+0.44\approx2.44\\2.44^{2.44}\approx8.8$
