The value of `7 log_(a) ""(16)/(15) + 5log_(a) ""(25)/(24) + 3 log_(a) ""(81)/(80)` is

Options

A. `log_(a)5`
B. `log_(a)3`
C. `log_(a)2`
D. None of these

Answer

To solve the problem \( 7 \log_a \left( \frac{16}{15} \right) + 5 \log_a \left( \frac{25}{24} \right) + 3 \log_a \left( \frac{81}{80} \right) \), we will follow these steps:

### Step 1: Rewrite the logarithmic expressions
We can express the fractions in terms of their prime factorization:
- \( 16 = 2^4 \)
- \( 15 = 3 \times 5 \)
- \( 25 = 5^2 \)
- \( 24 = 2^3 \times 3 \)
- \( 81 = 3^4 \)
- \( 80 = 2^4 \times 5 \)

Thus, we rewrite the logarithmic expressions as:
\[ 7 \log_a \left( \frac{2^4}{3 \times 5} \right) + 5 \log_a \left( \frac{5^2}{2^3 \times 3} \right) + 3 \log_a \left( \frac{3^4}{2^4 \times 5} \right) \]

### Step 2: Apply the logarithmic property \( x \log_a b = \log_a (b^x) \)
Using the property \( x \log_a b = \log_a (b^x) \), we can rewrite the expression:
\[ \log_a \left( \left( \frac{2^4}{3 \times 5} \right)^7 \right) + \log_a \left( \left( \frac{5^2}{2^3 \times 3} \right)^5 \right) + \log_a \left( \left( \frac{3^4}{2^4 \times 5} \right)^3 \right) \]

### Step 3: Combine the logarithms
Using the property \( \log_a b + \log_a c = \log_a (bc) \):
\[ \log_a \left( \left( \frac{2^4}{3 \times 5} \right)^7 \cdot \left( \frac{5^2}{2^3 \times 3} \right)^5 \cdot \left( \frac{3^4}{2^4 \times 5} \right)^3 \right) \]

### Step 4: Simplify the expression inside the logarithm
Calculating each term:
1. \( \left( \frac{2^4}{3 \times 5} \right)^7 = \frac{2^{28}}{3^7 \times 5^7} \)
2. \( \left( \frac{5^2}{2^3 \times 3} \right)^5 = \frac{5^{10}}{2^{15} \times 3^5} \)
3. \( \left( \frac{3^4}{2^4 \times 5} \right)^3 = \frac{3^{12}}{2^{12} \times 5^3} \)

Now combine these:
\[ \frac{2^{28}}{3^7 \times 5^7} \cdot \frac{5^{10}}{2^{15} \times 3^5} \cdot \frac{3^{12}}{2^{12} \times 5^3} \]

### Step 5: Combine the powers of each base
Combining the powers:
- For \(2\): \(28 - 15 - 12 = 1\)
- For \(3\): \(0 + 12 - 7 - 5 = 0\)
- For \(5\): \(10 - 7 - 3 = 0\)

Thus, we have:
\[ \frac{2^1 \cdot 3^0 \cdot 5^0}{1} = 2 \]

### Step 6: Final result
Now we can write:
\[ \log_a(2) \]

Thus, the final value of the expression is:
\[ \log_a(2) \]