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The given equation is:

\[
\frac{(27^2)^3 \cdot 27^5}{27^{7n}} = 27^2
\]

We need to simplify the left-hand side and solve for the whole number value of \( n \).

### Step 1: Simplify \((27^2)^3\)
\[
(27^2)^3 = 27^{2 \cdot 3} = 27^6
\]

So, the equation becomes:

\[
\frac{27^6 \cdot 27^5}{27^{7n}} = 27^2
\]

### Step 2: Combine the exponents in the numerator
\[
27^6 \cdot 27^5 = 27^{6 + 5} = 27^{11}
\]

Now, the equation looks like:

\[
\frac{27^{11}}{27^{7n}} = 27^2
\]

### Step 3: Apply the property of exponents \(\frac{a^m}{a^n} = a^{m-n}\)
\[
\frac{27^{11}}{27^{7n}} = 27^{11 - 7n}
\]

Now, we have:
\[
27^{11 - 7n} = 27^2
\]

### Step 4: Set the exponents equal to each other
Since the bases are the same (both are powers of 27), we can set the exponents equal to each other:

\[
11 - 7n = 2
\]

### Step 5: Solve for \( n \)
\[
11 - 7n = 2
\]
\[
11 - 2 = 7n
\]
\[
9 = 7n
\]
\[
n = \frac{9}{7}
\]

Since \( n = \frac{9}{7} \) is not a whole number, there seems to be an issue because the problem specifies a whole number. Let's review the calculation details again.

Do you want further clarification or details? Here's a set of questions that could help explore this:

1. How do we simplify powers of powers more effectively?
2. What exponent properties are applied in such problems?
3. Can we generalize the method for similar problems?
4. How do exponent equations with different bases affect solving methods?
5. What is the significance of setting exponents equal when the bases match?

**Tip:** Always check the equation step by step to avoid missing minor simplifications when dealing with powers.

What whole number value of n makes the equation \(\frac{(27^2)^3 \cdot 27^5}{27^{7n}} = 27^2\) true?

Learn how to find the whole number value of n that satisfies the exponential equation \(\frac{(27^2)^3 \cdot 27^5}{27^{7n}} = 27^2\). This problem covers exponent rules, including power of a power and the quotient of exponents, suitable for students in Grades 9-11.

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