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 equal to:

• A. $\log 2$
• B. $\log 3$
• C. $\log 5$
• D. $\log 6$

Answer 1

Answer: (A)

$\log 2$

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

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

$=\log [{\left(\frac{2^4}{3\times 5}\right)}^7.{\left(\frac{5^2}{2^3\times 3}\right)}^5.{\left(\frac{3^4}{2^4\times 5}\right)}^3]$

$=\log 2$