Question

Let $A=\frac{{\left({\log }_{2}3\right)}^3-{\left({\log }_{2}6\right)}^3-{\left({\log }_{2}12\right)}^3+{\left({\log }_{2}24\right)}^3}{6}$ then the value of $\left(2^A\right)$ is equal to

• A. 64
• B. 72
• C. 128
• D. 100

Answer 1

Answer: (B)

$\text{ Let }{\log }_{2}3=a$
$A=\frac{a^3-(a+1{)}^3-(a+2{)}^3+(a+3{)}^3}{6}$
$A=\frac{a^3-\left(a^3+1+3a^2+3a\right)-\left(a^3+8+6a^2+12a\right)+\left(a^3+27+9a^2+27a\right)}{6}$
$A=\frac{12a+18}{6}=2a+3$
$\therefore 2^A=2\cdot 8=\left(2^{{\log }_{2}9}\right)\times 8=9\times 8=72$