%5E2-2(4%5Ex-3*2%5Ex)-8-%3D0.jpeg "(4^x-3*2^x)^2-2(4^x-3*2^x)-8 =0")
Solving the Quadratic Equation: (4^x-32^x)^2-2(4^x-32^x)-8 = 0
In this article, we will discuss the solution to the quadratic equation (4^x-3*2^x)^2-2(4^x-3*2^x)-8 = 0. This equation involves exponential functions and can be challenging to solve. However, with the right approach, we can find the solution.
Simplifying the Equation
First, let's simplify the equation by expanding the exponentiation and combining like terms:
(4^x-3*2^x)^2 = (4^x)^2 - 2*4^x*3*2^x + (3*2^x)^2
= 4^(2x) - 24^x + 9*(2^x)^2
= 4^(2x) - 24^x + 9*2^(2x)
Now, substituting this expression back into the original equation, we get:
4^(2x) - 24^x + 9*2^(2x) - 2(4^x-3*2^x) - 8 = 0
Rearranging the Terms
Rearranging the terms, we get:
4^(2x) + 9*2^(2x) - 24^x - 8 - 2*4^x + 6*2^x = 0
Factoring the Expression
Notice that the expression can be factored as:
(2^x - 4)(2^x - 2)(4^x + 4) = 0
Solving for x
Now, we can set each factor equal to 0 and solve for x:
Case 1: 2^x - 4 = 0
2^x = 4
taking log base 2 on both sides
x = 2
Case 2: 2^x - 2 = 0
2^x = 2
taking log base 2 on both sides
x = 1
Case 3: 4^x + 4 = 0
4^x = -4
no real solution
Therefore, the solutions to the equation (4^x-3*2^x)^2-2(4^x-3*2^x)-8 = 0 are x = 1 and x = 2.
Conclusion
In this article, we have successfully solved the quadratic equation (4^x-3*2^x)^2-2(4^x-3*2^x)-8 = 0 using algebraic manipulations and exponentiation rules. The solutions to the equation are x = 1 and x = 2.
