Question

The value of ${\log }_2\left(20\right){\log }_2\left(80\right)-{\log }_2\left(5\right){\log }_2\left(320\right)$ is

• A. $5$
• B. $6$
• C. $7$
• D. $8$
• E. $10$

Answer 1

The correct option is D

$8$

Explanation for the correct answer:

Compute the required value:

Let $S={\log }_2\left(20\right){\log }_2\left(80\right)-{\log }_2\left(5\right){\log }_2\left(320\right)$

Write the numbers as the product of their prime factors

$\Rightarrow$ $S={\log }_2\left(2^2\times 5\right){\log }_2\left(2^4\times 5\right)-{\log }_2\left(5\right){\log }_2\left(2^6\times 5\right)$

$\Rightarrow$ $S=\left({\log }_2\left(2^2\right)+{\log }_2\left(5\right)\right)\left({\log }_2\left(2^4\right)+{\log }_2\left(5\right)\right)-{\log }_2\left(5\right)\left({\log }_2\left(2^6\right)+{\log }_2\left(5\right)\right)$ $\left[\because \log \left(ab\right)=\log \left(a\right)+\log \left(b\right)\right]$

$\Rightarrow$ $S=\left(2{\log }_2\left(2\right)+{\log }_2\left(5\right)\right)\left(4{\log }_2\left(2\right)+{\log }_2\left(5\right)\right)-{\log }_2\left(5\right)\left(6{\log }_2\left(2\right)+{\log }_2\left(5\right)\right)$ $\left[\because \log \left(a^m\right)=m\log \left(a\right)\right]$

$\Rightarrow$ $S=8{\left({\log }_2\left(2\right)\right)}^2+6{\log }_2\left(2\right){\log }_2\left(5\right)+{\left({\log }_2\left(5\right)\right)}^2-6{\log }_2\left(2\right){\log }_2\left(5\right)-{\left({\log }_2\left(5\right)\right)}^2$ $\left[\because {\log }_2\left(2\right)=1\right]$

$\Rightarrow$ $S=8\left(1^2\right)$

$\Rightarrow$ $S=8$

Hence, the value of ${\log }_2\left(20\right){\log }_2\left(80\right)-{\log }_2\left(5\right){\log }_2\left(320\right)$ is $8$.

Hence, option $\left(D\right)$ is the correct answer.