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(4^x-3*2^x)^2-2(4^x-3*2^x)-8 =0

3 min read Jun 16, 2024
(4^x-3*2^x)^2-2(4^x-3*2^x)-8 =0

Solving the Equation (4^x - 32^x)^2 - 2(4^x - 32^x) - 8 = 0

This equation may seem intimidating at first glance, but we can solve it by using a clever substitution and applying our knowledge of quadratic equations. Here's how:

1. Recognizing the Pattern

Observe that the expression (4^x - 3*2^x) appears twice in the equation. This suggests a substitution to simplify the equation.

2. The Substitution

Let's substitute y = 4^x - 3*2^x. Now, the equation becomes:

y^2 - 2y - 8 = 0

3. Solving the Quadratic Equation

This is a standard quadratic equation that can be solved by factoring or using the quadratic formula.

  • Factoring: (y - 4)(y + 2) = 0 This gives us two solutions: y = 4 or y = -2

  • Quadratic Formula: y = (-b ± √(b^2 - 4ac)) / 2a Where a = 1, b = -2, and c = -8. This also leads to the same solutions: y = 4 or y = -2

4. Back-Substituting

Now, we need to substitute back y = 4^x - 3*2^x to find the values of x.

  • For y = 4: 4^x - 32^x = 4 (2^2)^x - 32^x - 4 = 0 Let z = 2^x, then the equation becomes: z^2 - 3z - 4 = 0 Factoring: (z - 4)(z + 1) = 0 This gives us z = 4 or z = -1. Since 2^x is always positive, we discard z = -1. Therefore, 2^x = 4, which means x = 2.

  • For y = -2: 4^x - 32^x = -2 (2^2)^x - 32^x + 2 = 0 Let z = 2^x, then the equation becomes: z^2 - 3z + 2 = 0 Factoring: (z - 2)(z - 1) = 0 This gives us z = 2 or z = 1. Therefore, 2^x = 2 (x = 1) or 2^x = 1 (x = 0).

5. The Solutions

We have found three solutions for the original equation:

  • x = 0
  • x = 1
  • x = 2

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