Question

• A. 7
• B. 17/3
• C. 11
• D. 16/3

Answer 1

Answer:

B. 17/3

$2^{3k-5}=4096$

$\frac{2^{3k}}{2^5}=2^{12}$

$2^{3k}=2^{12}\times 2^5$

$2^{3k}=2^{17}$

$3k=17$

$k=17/3$