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(sqrt(32))^(x)- 2^(y+1)=1 16^(4-(x)/(2))-8^(y)=0

3 min read Jun 16, 2024
(sqrt(32))^(x)- 2^(y+1)=1 16^(4-(x)/(2))-8^(y)=0

Solving a System of Equations with Exponential Terms

This article will explore the process of solving the following system of equations:

(1) √(32)^(x) - 2^(y+1) = 1 (2) 16^(4-(x)/(2)) - 8^(y) = 0

Understanding the Equations

Both equations involve exponential terms. Let's break them down:

  • Equation (1):

    • √(32) can be simplified to 2^(5/2).
    • Therefore, the equation becomes: (2^(5/2))^(x) - 2^(y+1) = 1
    • Using the rule (a^m)^n = a^(m*n), we get: 2^(5x/2) - 2^(y+1) = 1

  • Equation (2):

    • 16 and 8 can be expressed as powers of 2: 16 = 2^4 and 8 = 2^3.
    • The equation becomes: (2^4)^(4-(x/2)) - (2^3)^y = 0
    • Simplifying: 2^(16-2x) - 2^(3y) = 0

Solving the System

To solve the system, we will use substitution:

  1. Solve for one variable in terms of the other: Let's solve equation (1) for 2^(y+1):

    • 2^(y+1) = 2^(5x/2) - 1

  2. Substitute: Substitute this expression for 2^(y+1) into equation (2):

    • 2^(16-2x) - (2^(5x/2) - 1)^3 = 0

  3. Simplify and Solve for x:

    • Expanding the cube, we get a polynomial equation in terms of 2^(5x/2). This equation can be solved using various methods, including factoring, the quadratic formula, or numerical methods.

  4. Substitute x to find y: Once we find the value(s) of x, we can substitute them back into the equation 2^(y+1) = 2^(5x/2) - 1 to solve for y.

Important Considerations

  • Domain: Ensure that the solutions you obtain for x and y are within the domain of the original equations.
  • Extraneous Solutions: It's important to check if any solutions obtained are extraneous.

Conclusion

Solving the system of equations involving exponential terms requires a combination of simplifying, substitution, and solving for the variables. The process involves careful manipulation of the exponents and may require advanced algebraic techniques to find the solutions.

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