`7log((16)/(15))+5log((25)/(24))+3log((81)/(80))` is equal to

Options

A. `0`
B. `1`
C. `log 2`
D. `log 3`

Answer

To solve the expression \( 7\log\left(\frac{16}{15}\right) + 5\log\left(\frac{25}{24}\right) + 3\log\left(\frac{81}{80}\right) \), we will use the properties of logarithms.

### Step 1: Apply the power rule of logarithms
Using the property \( m\log(a) = \log(a^m) \), we can rewrite the expression:

\[ 7\log\left(\frac{16}{15}\right) = \log\left(\left(\frac{16}{15}\right)^7\right) \]
\[ 5\log\left(\frac{25}{24}\right) = \log\left(\left(\frac{25}{24}\right)^5\right) \]
\[ 3\log\left(\frac{81}{80}\right) = \log\left(\left(\frac{81}{80}\right)^3\right) \]

So, the expression becomes:

\[ \log\left(\left(\frac{16}{15}\right)^7\right) + \log\left(\left(\frac{25}{24}\right)^5\right) + \log\left(\left(\frac{81}{80}\right)^3\right) \]

### Step 2: Use the property of logarithms to combine
Using the property \( \log(a) + \log(b) = \log(ab) \), we can combine the logarithms:

\[ \log\left(\left(\frac{16}{15}\right)^7 \cdot \left(\frac{25}{24}\right)^5 \cdot \left(\frac{81}{80}\right)^3\right) \]

### Step 3: Calculate the expression inside the logarithm
Now we need to simplify the expression inside the logarithm:

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

Combining these, we have:

\[ \log\left(\frac{16^7 \cdot 25^5 \cdot 81^3}{15^7 \cdot 24^5 \cdot 80^3}\right) \]

### Step 4: Substitute the powers
Now we can substitute the values:

- \( 16 = 2^4 \) so \( 16^7 = (2^4)^7 = 2^{28} \)
- \( 25 = 5^2 \) so \( 25^5 = (5^2)^5 = 5^{10} \)
- \( 81 = 3^4 \) so \( 81^3 = (3^4)^3 = 3^{12} \)

Thus, the numerator becomes:

\[ 2^{28} \cdot 5^{10} \cdot 3^{12} \]

For the denominator:

- \( 15 = 3 \cdot 5 \) so \( 15^7 = (3 \cdot 5)^7 = 3^7 \cdot 5^7 \)
- \( 24 = 2^3 \cdot 3 \) so \( 24^5 = (2^3 \cdot 3)^5 = 2^{15} \cdot 3^5 \)
- \( 80 = 2^4 \cdot 5 \) so \( 80^3 = (2^4 \cdot 5)^3 = 2^{12} \cdot 5^3 \)

Thus, the denominator becomes:

\[ 3^7 \cdot 5^7 \cdot 2^{15} \cdot 3^5 \cdot 2^{12} \cdot 5^3 = 2^{27} \cdot 3^{12} \cdot 5^{10} \]

### Step 5: Combine and simplify
Now we can combine the numerator and denominator:

\[ \log\left(\frac{2^{28} \cdot 5^{10} \cdot 3^{12}}{2^{27} \cdot 3^{12} \cdot 5^{10}}\right) = \log\left(\frac{2^{28}}{2^{27}}\right) = \log(2) = 1 \]

### Final Result
Thus, the final result is:

\[ \log(2) \]