We need to solve the equation:

$\left(\frac{1}{9} \times 27^x\right)^2 = \sqrt{3^{4x}}$

Step 1: Simplifying the Left-Hand Side

Start with the expression:

$\left(\frac{1}{9} \times 27^x\right)^2$

First, express 9 and 27 in terms of powers of 3:

$9 = 3^2 \quad \text{and} \quad 27 = 3^3$

So the expression becomes:

$\left(\frac{1}{3^2} \times (3^3)^x\right)^2$

This simplifies to:

$\left(\frac{1}{3^2} \times 3^{3x}\right)^2$

Which further simplifies to:

$\left(3^{3x - 2}\right)^2$

Applying the power rule $(a^m)^n = a^{mn}$, we get:

$3^{2(3x - 2)} = 3^{6x - 4}$

Step 2: Simplifying the Right-Hand Side

Now, simplify the right-hand side:

$\sqrt{3^{4x}}$

The square root of $3^{4x}$ is:

$3^{\frac{4x}{2}} = 3^{2x}$

Step 3: Setting the Expressions Equal

We now have:

$3^{6x - 4} = 3^{2x}$

Step 4: Equating the Exponents

Since the bases are the same, set the exponents equal:

$6x - 4 = 2x$

Step 5: Solving for $x$

Subtract $2x$ from both sides:

$4x - 4 = 0$

Add 4 to both sides:

$4x = 4$

Divide both sides by 4:

$x = 1$

Final Answer:

The solution is $x = 1$.

Would you like more details or have any questions?

Related Questions:

1. How would the solution change if the exponent on the left-hand side was different?
2. What if the right-hand side involved a cube root instead of a square root?
3. How do power rules apply when working with fractional exponents?
4. Can logarithms be used to solve this equation more directly?
5. How do you handle equations where the bases differ on each side?

Tip:

When solving exponential equations, always try to express all terms using the same base for easier comparison and simplification.