Question

If $E=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.......\frac{30}{62}.\frac{31}{64}=8^x$ , then the value of $x$ is:

• A. $-7$
• B. $-9$
• C. $-10$
• D. $-12$

Answer 1

The correct option is D $-12$

$E=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}...\frac{30}{62}.\frac{31}{64}$

$=\frac{31!}{4\times 6\times 8\times \mathrm{10...}\times 64}$

$=\frac{31!}{2^{31}.(1\times 2\times 3\times \mathrm{4...}\times 32}$

$=\frac{31!}{2^{31}.32!}$

$=\frac{1}{2^{31}.32}$

$=\frac{1}{2^{36}}$

$=\frac{1}{(2^3{)}^{12}}$

$=\frac{1}{8^{12}}$

$=8^{-12}$

Hence $x=-12$.