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+3 a^2+3 a\right)-\left(a^3+8+6 a^2+12 a\right)+\left(a^3+27+9 a^2+27 a\right)}{6} $
$A=\frac{12 a+18}{6}=2 a+3 $
$\therefore 2^A=2 \cdot 8=\left(2^{\log _2 9}\right) \times 8=9 \times 8=72$