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 = 0y = 1/4
y - 9 = 0y = 9
Since y = 2^x, this means:
2^x = 1/4x = -2
2^x = 9x = 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.
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.