Math Problem Statement

Solve the equation: 2^(4x-1) = (1/4)^(2-x)

Solution

The equation given in the image is:

24x1=(14)2x2^{4x - 1} = \left( \frac{1}{4} \right)^{2 - x}

Let's solve this step-by-step.

Step 1: Rewrite the right-hand side

14=22\frac{1}{4} = 2^{-2} So the equation becomes: 24x1=(22)2x2^{4x - 1} = \left( 2^{-2} \right)^{2 - x}

Step 2: Simplify the right-hand side

Using the power of a power rule (am)n=amn(a^m)^n = a^{m \cdot n}, we get: 24x1=22(2x)=24+2x2^{4x - 1} = 2^{-2(2 - x)} = 2^{-4 + 2x}

Step 3: Equate the exponents

Since the bases are the same (base 2), we can equate the exponents: 4x1=4+2x4x - 1 = -4 + 2x

Step 4: Solve for xx

Subtract 2x2x from both sides: 4x2x1=44x - 2x - 1 = -4 2x1=42x - 1 = -4

Add 1 to both sides: 2x=32x = -3

Divide by 2: x=32x = -\frac{3}{2}

Final Answer:

The solution is x=1.5x = -1.5.

Would you like more details on any part of this solution?

Here are 5 related questions for further exploration:

  1. What is the general rule for equating exponents when bases are the same?
  2. How do you solve exponential equations when the bases are different?
  3. What other methods can be used to solve equations like this one?
  4. What are some common applications of exponential equations in real-world problems?
  5. How can we check the solution of an exponential equation?

Tip: Always remember to check your solution by substituting it back into the original equation!

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Math Problem Analysis

Mathematical Concepts

Exponential Equations
Algebra

Formulas

(a^m)^n = a^(m * n)

Theorems

Equating exponents with the same base

Suitable Grade Level

Grades 10-12