To solve the equation:

$\left( \sqrt{27} \right)^{2x-5} = \frac{1}{3^6}$

Step 1: Simplify the square root

The square root of $27$ can be written as: $\sqrt{27} = 27^{1/2} = (3^3)^{1/2} = 3^{3/2}.$

Thus, the base becomes $3^{3/2}$.

Step 2: Rewrite the left-hand side

$\left( \sqrt{27} \right)^{2x-5} = \left( 3^{3/2} \right)^{2x-5}.$

Using the power rule $(a^m)^n = a^{m \cdot n}$: $\left( 3^{3/2} \right)^{2x-5} = 3^{(3/2)(2x-5)} = 3^{3x - 15/2}.$

Step 3: Rewrite the right-hand side

$\frac{1}{3^6} = 3^{-6}.$

Step 4: Equate the exponents

Since the bases are the same, equate the exponents: $3x - \frac{15}{2} = -6.$

Step 5: Solve for $x$

1. Add $\frac{15}{2}$ to both sides: $3x = -6 + \frac{15}{2}.$

2. Rewrite $-6$ as $-\frac{12}{2}$ for a common denominator: $3x = -\frac{12}{2} + \frac{15}{2}.$

3. Simplify: $3x = \frac{3}{2}.$

4. Divide both sides by 3: $x = \frac{3}{2} \div 3 = \frac{3}{2} \cdot \frac{1}{3} = \frac{1}{2}.$

Final Answer:

The value of $x$ is: $\boxed{\frac{1}{2}}.$

Let me know if you have any questions or want further clarification!

Here are 5 related questions to practice:

1. How do you simplify $(a^m)^n$ when dealing with exponents?
2. What is the square root of $81$ written as a power of 3?
3. How would you rewrite $\frac{1}{x^n}$ using a negative exponent?
4. What are the properties of exponents that allow simplification of fractional powers?
5. Can you solve $(\sqrt{16})^{2x+1} = 4^{-2}$ for $x$?

Tip: Always express roots and fractional bases in terms of their prime factors when solving exponent equations.