We are given the equation:

$\sqrt{3^{2a + b}} = \frac{27^a}{3^b}$

Step 1: Simplify the square root

First, express the square root in exponent form:

$\sqrt{3^{2a + b}} = \left(3^{2a + b}\right)^{\frac{1}{2}} = 3^{\frac{2a + b}{2}}$

Step 2: Simplify the right-hand side

Next, express $27^a$ in terms of base 3. Since $27 = 3^3$, we have:

$27^a = (3^3)^a = 3^{3a}$

Thus, the right-hand side of the equation becomes:

$\frac{27^a}{3^b} = \frac{3^{3a}}{3^b} = 3^{3a - b}$

Step 3: Set the exponents equal

Now, set the exponents of the base 3 equal to each other:

$\frac{2a + b}{2} = 3a - b$

Step 4: Solve for $a$

Multiply both sides of the equation by 2 to eliminate the fraction:

$2a + b = 2(3a - b)$

Expand the right-hand side:

$2a + b = 6a - 2b$

Now, move all terms involving $a$ to one side and all terms involving $b$ to the other side:

$2a - 6a = -2b - b$

Simplify:

$-4a = -3b$

Finally, divide by $-4$:

$a = \frac{3b}{4}$

Final answer:

$a = \frac{3b}{4}$

Would you like further clarification or additional details?

Here are 5 related questions to explore:

1. How would the solution change if the base of the powers was different?
2. What if the equation involved cube roots instead of square roots?
3. How can you solve similar equations involving different variables or constants?
4. What happens if you square both sides of the original equation instead?
5. How does logarithmic manipulation help in solving exponential equations?

Tip: When dealing with exponents and roots, always try to express terms with the same base for easier comparison.