To solve the equation 16^(x+1/4) - 41*4^(x-1) + 9 = 0, we can first rewrite the terms in terms of their base as follows:

16^(x+1/4) = (2^4)^(x+1/4) = 2^(4(x+1/4)) = 2^(4x+1)

414^(x-1) = 41(2^2)^(x-1) = 412^(2(x-1)) = 412^(2x-2)

So the equation becomes:

2^(4x+1) - 41*2^(2x-2) + 9 = 0

Now, we can let y = 2^x, which means 2^x = y. Then the equation becomes:

2^(4x+1) = 2^(2) * 2^(2x) = 4y^2

Substituting into the equation:

4y^2 - 41y + 9 = 0

This is a quadratic equation that can be factored as:

(4y - 1)(y - 9) = 0

Setting each factor to zero:

4y - 1 = 0
y = 1/4

y - 9 = 0
y = 9

Since y = 2^x, this means:

2^x = 1/4
x = -2

2^x = 9
x = log(9)/log(2) = 3.169925

Therefore, the solutions to the equation 16^(x+1/4) - 41*4^(x-1) + 9 = 0 are x = -2 and x ≈ 3.169925.