Question

The value of $7 \log\left(\frac{16}{15} \right) +5 \log\left(\frac{25}{24}\right) + 3 \log\left(\frac{81}{80}\right)$ is equla to

• A. log 2
• B. 3
• C. 5
• D. 7

Answer 1

Answer: (A)

log 2

Answer 2

$\therefore 7 \log \left(\frac{16}{15}\right)+5 \log \left(\frac{25}{24}\right)+3 \log \left(\frac{81}{80}\right)$ $=\log \left[\left(\frac{16}{15}\right)^{7} \cdot\left(\frac{25}{24}\right)^{5} \cdot\left(\frac{81}{80}\right)^{3}\right]$ $=\log 2$