https://file.aiquickdraw.com/mathbot/file/1726143593145wpem2w2q.jpg

The equation given in the image is:

\[
2^{4x - 1} = \left( \frac{1}{4} \right)^{2 - x}
\]

Let's solve this step-by-step.

### Step 1: Rewrite the right-hand side
\[
\frac{1}{4} = 2^{-2}
\]
So the equation becomes:
\[
2^{4x - 1} = \left( 2^{-2} \right)^{2 - x}
\]

### Step 2: Simplify the right-hand side
Using the power of a power rule \((a^m)^n = a^{m \cdot n}\), we get:
\[
2^{4x - 1} = 2^{-2(2 - x)} = 2^{-4 + 2x}
\]

### Step 3: Equate the exponents
Since the bases are the same (base 2), we can equate the exponents:
\[
4x - 1 = -4 + 2x
\]

### Step 4: Solve for \(x\)
Subtract \(2x\) from both sides:
\[
4x - 2x - 1 = -4
\]
\[
2x - 1 = -4
\]

Add 1 to both sides:
\[
2x = -3
\]

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

### Final Answer:
The solution is \(x = -1.5\).

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!

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

This problem demonstrates how to solve the exponential equation 2^(4x-1) = (1/4)^(2-x) using properties of exponents and algebra. A step-by-step solution is provided for students in Grades 10-12.

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