# How do you solve  7^x= 7^(2^(x^-3))?

This equality can only be solved with approximations, and the result is something like $x = 1.33670973494950 \ldots$
Since the bases are the same (7 in both sides), we have that ${7}^{x} = {7}^{{2}^{{x}^{- 3}}}$ if and only if the exponents are equal, i.e. if and only if
$x = {2}^{{x}^{-} 3}$