There is no algebraic way to solve an equation when the unknown
appears both as part of an exponent and elsewhere in the same
equation as not part of an exponent.
However by inspection of the left side, the powers of 4 are
4,16,64, etc. Also by inspection of the right side, the multiples
of 8 are 8,16,24, etc., so we notice that x=2 is a solution.
But there is another solution, which can be approximated with a
graphing calculator or by iterative methods, x = 0.1549534662,
correct to ten decimal places.
Edwin
