√2^n = 1024, then find the value of 3^2(n/4 - 4)

We are given the equation:

\[
\sqrt{2^n} = 1024
\]

### Step 1: Simplify the square root equation
We know that \( \sqrt{2^n} = 2^{n/2} \). So the equation becomes:

\[
2^{n/2} = 1024
\]

### Step 2: Express 1024 as a power of 2
We can rewrite \( 1024 \) as \( 2^{10} \) because \( 2^{10} = 1024 \). This gives us:

\[
2^{n/2} = 2^{10}
\]

### Step 3: Solve for \( n \)
Since the bases are the same, we can equate the exponents:

\[
\frac{n}{2} = 10
\]

Multiplying both sides by 2:

\[
n = 20
\]

### Step 4: Find the value of \( 3^{2(n/4 - 4)} \)
Now that we know \( n = 20 \), substitute \( n \) into the expression \( 3^{2(n/4 - 4)} \):

\[
n/4 = \frac{20}{4} = 5
\]

So the expression becomes:

\[
3^{2(5 - 4)} = 3^{2 \times 1} = 3^2 = 9
\]

Thus, the value of \( 3^{2(n/4 - 4)} \) is \( \boxed{9} \).

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you convert a square root equation to an exponent form?
2. Can you express other powers of 2, such as 512 or 2048, and solve similar equations?
3. How would this problem change if \( \sqrt{3^n} = 1024 \)?
4. What are the properties of exponents that allow you to equate powers?
5. How can logarithms be used to solve for \( n \) in this problem?

**Tip:** Always break down powers and roots into simpler components (like expressing numbers as powers of 2 or 3) to make solving easier.

Learn how to solve the equation √2^n = 1024 and find the value of 3^2(n/4 - 4). This step-by-step guide covers key concepts such as simplifying square roots, powers of 2, and solving exponent equations. Suitable for high school students (Grades 9-12).

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