If `9^(3)xx 27^(2)xx 81^(2)` is equal to `3^(2k)`, then find k.

Answer

To solve the problem \(9^3 \times 27^2 \times 81^2 = 3^{2k}\), we will first express each base in terms of powers of 3.

### Step 1: Rewrite the bases in terms of powers of 3
- \(9\) can be written as \(3^2\).
- \(27\) can be written as \(3^3\).
- \(81\) can be written as \(3^4\).

### Step 2: Substitute the rewritten bases into the equation
Now, substituting these values into the original expression:
\[ 9^3 = (3^2)^3 = 3^{2 \times 3} = 3^6 \]
\[ 27^2 = (3^3)^2 = 3^{3 \times 2} = 3^6 \]
\[ 81^2 = (3^4)^2 = 3^{4 \times 2} = 3^8 \]

### Step 3: Combine the powers of 3
Now, we can combine these results:
\[ 9^3 \times 27^2 \times 81^2 = 3^6 \times 3^6 \times 3^8 \]
Using the property of exponents that states \(a^m \times a^n = a^{m+n}\):
\[ 3^6 \times 3^6 \times 3^8 = 3^{6 + 6 + 8} = 3^{20} \]

### Step 4: Set the equation equal to \(3^{2k}\)
Now we have:
\[ 3^{20} = 3^{2k} \]

### Step 5: Equate the exponents
Since the bases are the same, we can set the exponents equal to each other:
\[ 20 = 2k \]

### Step 6: Solve for \(k\)
To find \(k\), divide both sides by 2:
\[ k = \frac{20}{2} = 10 \]

Thus, the value of \(k\) is \(10\).

### Final Answer:
\(k = 10\)

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