How much does this speed up brute force search, though? You have to actually compute the exponent and digit sums. If I iterate over the powers of N from N^I to N^J, I do something like O((J-I)^2*log N) work doing multiplications the naive way.

Rewrite

    log(x^y)*4.5 = x
    y*log(x)*4.5 = x
    y = x/(4.5*log x)

So picking a random number like 234, we get y=21.94819 and the digit sum of 234^22 == 234. In this case we actually just had to compute one exponent and one digit sum.

Hint: try checking other numbers.