If `7"log"_(a)(16)/(15)+5"log"_(a)(25)/(24)+3"log"_(a)(81)/(80)=8`, then `a=`

Options

A. `2^(1//8)`
B. `(10)^(1//8)`
C. `(30)^(1//8)`
D. 1

Answer

To solve the equation

\[ 7 \log_a \left(\frac{16}{15}\right) + 5 \log_a \left(\frac{25}{24}\right) + 3 \log_a \left(\frac{81}{80}\right) = 8, \]

we will use properties of logarithms.

### Step 1: Apply the logarithmic identity

We know that \( \log_a \left(\frac{x}{y}\right) = \log_a x - \log_a y \). We will apply this property to each term in the equation.

\[ 7 \left( \log_a 16 - \log_a 15 \right) + 5 \left( \log_a 25 - \log_a 24 \right) + 3 \left( \log_a 81 - \log_a 80 \right) = 8. \]

### Step 2: Expand the equation

Expanding the equation gives us:

\[ 7 \log_a 16 - 7 \log_a 15 + 5 \log_a 25 - 5 \log_a 24 + 3 \log_a 81 - 3 \log_a 80 = 8. \]

### Step 3: Combine like terms

Now, we can group the logarithmic terms:

\[ (7 \log_a 16 + 5 \log_a 25 + 3 \log_a 81) - (7 \log_a 15 + 5 \log_a 24 + 3 \log_a 80) = 8. \]

### Step 4: Substitute logarithmic values

Next, we will substitute the logarithmic values using their prime factorization:

- \( 16 = 2^4 \) so \( \log_a 16 = 4 \log_a 2 \)
- \( 25 = 5^2 \) so \( \log_a 25 = 2 \log_a 5 \)
- \( 81 = 3^4 \) so \( \log_a 81 = 4 \log_a 3 \)
- \( 15 = 3 \times 5 \) so \( \log_a 15 = \log_a 3 + \log_a 5 \)
- \( 24 = 2^3 \times 3 \) so \( \log_a 24 = 3 \log_a 2 + \log_a 3 \)
- \( 80 = 2^4 \times 5 \) so \( \log_a 80 = 4 \log_a 2 + \log_a 5 \)

Substituting these into the equation gives:

\[ 7(4 \log_a 2) + 5(2 \log_a 5) + 3(4 \log_a 3) - \left[7(\log_a 3 + \log_a 5) + 5(3 \log_a 2 + \log_a 3) + 3(4 \log_a 2 + \log_a 5)\right] = 8. \]

### Step 5: Simplify the equation

This simplifies to:

\[ 28 \log_a 2 + 10 \log_a 5 + 12 \log_a 3 - \left[7 \log_a 3 + 7 \log_a 5 + 15 \log_a 2 + 3 \log_a 5 + 12 \log_a 2\right] = 8. \]

Combining like terms gives:

\[ (28 \log_a 2 - 15 \log_a 2 - 12 \log_a 2) + (10 \log_a 5 - 7 \log_a 5 - 3 \log_a 5) + (12 \log_a 3 - 7 \log_a 3) = 8. \]

This simplifies to:

\[ 1 \log_a 2 + 0 \log_a 5 + 5 \log_a 3 = 8. \]

### Step 6: Solve for \( a \)

This means:

\[ \log_a 2 + 5 \log_a 3 = 8. \]

Rearranging gives:

\[ \log_a 2 = 8 - 5 \log_a 3. \]

Using the change of base formula, we can express \( a \) in terms of \( 2 \) and \( 3 \):

\[ a^{8 - 5 \log_a 3} = 2. \]

Taking antilogarithm gives:

\[ a^{8} = 2 \cdot a^{5 \log_a 3}. \]

Thus, we can solve for \( a \).

### Final Step: Find the value of \( a \)

From the equation, we can deduce that:

\[ a^8 = 2 \cdot 3^5. \]

Taking the eighth root gives:

\[ a = \sqrt[8]{2 \cdot 243} = \sqrt[8]{486}. \]

### Conclusion

Thus, the value of \( a \) is:

\[ a = 2^{1/8} \cdot 3^{5/8}. \]