`3^(6 . 5) *27^(-5) xx 9^(2) = 3^(?)`

Options

A. a. 5
B. b. 4.5
C. c. 7
D. d. 6.5

Answer

To solve the equation \(3^{6.5} \cdot 27^{-5} \cdot 9^{2} = 3^{?}\), we will express all terms with the same base, which is 3.

### Step-by-step Solution:

1. **Rewrite \(27\) and \(9\) in terms of base \(3\)**:
- We know that \(27 = 3^3\) and \(9 = 3^2\).
- Therefore, we can rewrite the equation as:
\[ 3^{6.5} \cdot (3^3)^{-5} \cdot (3^2)^{2} \]

2. **Apply the power of a power rule**:
- Using the rule \((a^m)^n = a^{m \cdot n}\):
- \((3^3)^{-5} = 3^{-15}\)
- \((3^2)^{2} = 3^{4}\)
- So, the equation now becomes:
\[ 3^{6.5} \cdot 3^{-15} \cdot 3^{4} \]

3. **Combine the exponents**:
- When multiplying with the same base, we add the exponents:
\[ 3^{6.5 + (-15) + 4} \]

4. **Calculate the exponent**:
- Now, we simplify the exponent:
\[ 6.5 - 15 + 4 = 6.5 + 4 - 15 = 10.5 - 15 = -4.5 \]
- Thus, we have:
\[ 3^{-4.5} \]

5. **Set the exponent equal to the question mark**:
- From the equation \(3^{6.5} \cdot 27^{-5} \cdot 9^{2} = 3^{?}\), we find that:
\[ ? = -4.5 \]

### Final Answer:
The value of the question mark is \(-4.5\).

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