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)

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