Solving the Equation (8^x1)^2  18(8^x1) + 32 = 0
This equation might seem intimidating at first, but we can solve it using a simple substitution and factoring. Here's how:
1. Substitution
Let's make the substitution y = 8^x  1. This allows us to rewrite the equation in a much simpler form:
y^2  18y + 32 = 0
Now we have a quadratic equation in terms of 'y'.
2. Factoring the Quadratic
This quadratic equation can be easily factored:
(y  16)(y  2) = 0
This gives us two possible solutions for 'y':
 y = 16
 y = 2
3. Solving for x
Now we need to substitute back our original expression for 'y' and solve for 'x':

Case 1: y = 16
8^x  1 = 16 8^x = 17 x = log8(17)

Case 2: y = 2
8^x  1 = 2 8^x = 3 x = log8(3)
4. The Solutions
Therefore, the solutions to the equation (8^x1)^2  18(8^x1) + 32 = 0 are:
 x = log8(17)
 x = log8(3)
These solutions can be expressed in terms of natural logarithms if desired.
Conclusion
By using substitution and factoring, we successfully solved the equation and found its two solutions. This approach can be applied to similar equations involving exponential expressions.