Question

The value of $7{\log }_2\left(\frac{16}{15}\right)+5{\log }_2\left(\frac{25}{24}\right)+3{\log }_2\left(\frac{81}{80}\right)$ is

• A. $1$
• B. ${\log }_2\left(105\right)$
• C. ${\log }_2\left(\frac{9}{80}\right)$
• D. ${\log }_2\left(\frac{8}{9}\right)$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Given, $7{\log }_2\left(\frac{16}{15}\right)+5{\log }_2\left(\frac{25}{24}\right)+3{\log }_2\left(\frac{81}{80}\right)$

$$\begin{gathered} =7{\log }_2\left(16\right)-7{\log }_2\left(15\right)+5{\log }_2\left(25\right)-5{\log }_2\left(24\right)+3{\log }_2\left(81\right)-3{\log }_2\left(80\right) \\ =28{\log }_{2}2-7{\log }_{2}5-7{\log }_{2}3+10{\log }_{2}5-5{\log }_{2}8-5{\log }_{2}3+12{\log }_{2}3-3{\log }_{2}16-3{\log }_{2}5\left[\because {\log }_a{\left(b\right)}^n={\mathrm{nlog}}_a\left(b\right)\right] \\ =28{\log }_{2}2-15{\log }_{2}2-12{\log }_{2}2-7{\log }_{2}5+10{\log }_{2}5-3{\log }_{2}5-7{\log }_{2}3-5{\log }_{2}3+12{\log }_{2}3\left[\because {\log }_a\left(a\right)=1\right] \\ ={\log }_{2}2+0+0 \\ =1 \end{gathered}$$

Hence, the correct option is option (A), $1.$