Math Problem Statement

Solve the equation 2^(2x) - 6 * 2^x + 8 = 0 for x.

Solution

The equation provided is:

22x62x+8=02^{2x} - 6 \cdot 2^x + 8 = 0

This can be solved by using a substitution to simplify the equation. Let's follow these steps:

Step 1: Substitution

Let y=2xy = 2^x. This gives us the following substitution: 22x=(2x)2=y22^{2x} = (2^x)^2 = y^2 So the original equation becomes: y26y+8=0y^2 - 6y + 8 = 0

Step 2: Solve the quadratic equation

Now we solve the quadratic equation: y26y+8=0y^2 - 6y + 8 = 0 We can factor this as: (y4)(y2)=0(y - 4)(y - 2) = 0

Step 3: Solve for yy

From the factored equation, we have two possible solutions: y=4ory=2y = 4 \quad \text{or} \quad y = 2

Step 4: Solve for xx

Recall that y=2xy = 2^x, so now we solve for xx in each case:

  • If y=4y = 4, then 2x=42^x = 4. Since 4=224 = 2^2, we have: x=2x = 2

  • If y=2y = 2, then 2x=22^x = 2. This gives: x=1x = 1

Final Solutions:

The solutions for xx are: x=2orx=1x = 2 \quad \text{or} \quad x = 1

Let me know if you need more details or clarifications! Here are some related questions:

  1. How do we solve other types of exponential equations?
  2. What are common substitution methods for exponential equations?
  3. How can we check if the solutions to an equation are correct?
  4. How can we use logarithms to solve exponential equations?
  5. What are the properties of exponents that simplify solving exponential equations?

Tip: When faced with exponential equations, try using substitution to reduce the complexity of the problem before solving.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Exponential Equations
Quadratic Equations

Formulas

Quadratic equation formula ax^2 + bx + c = 0

Theorems

Quadratic formula

Suitable Grade Level

Grades 9-11