The value of frac{9^{2} times 18^{4}}{3^{16}} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given expression is $\frac{9^{2} 	imes 18^{4}}{3^{16}}$.Expressing the bases in terms of prime factors:$9 = 3^{2}$ and $18 = 2 	imes 3^{2}$.Subs

Vedclass · Accepted Answer

$\frac{16}{81}$

Vedclass · Answer

(C) Given expression is $\frac{9^{2} 	imes 18^{4}}{3^{16}}$.Expressing the bases in terms of prime factors:$9 = 3^{2}$ and $18 = 2 	imes 3^{2}$.Substituting these into the expression:$= \frac{(3^{2})^{2} 	imes (2 	imes 3^{2})^{4}}{3^{16}}$$= \frac{3^{4} 	imes 2^{4} 	imes (3^{2})^{4}}{3^{16}}$$= \frac{3^{4} 	imes 2^{4} 	imes 3^{8}}{3^{16}}$$= \frac{3^{4+8} 	imes 2^{4}}{3^{16}}$$= \frac{3^{12} 	imes 2^{4}}{3^{16}}$$= \frac{2^{4}}{3^{16-12}}$$= \frac{16}{3^{4}}$$= \frac{16}{81}$.

The value of $\frac{9^{2} \times 18^{4}}{3^{16}}$ is

Similar Questions

Determine the value of $x$. If $x = \frac{(943 + 864)^{2} - (943 - 864)^{2}}{(1886 \times 1728)}$

If $x>1$ and $x^{2}+\frac{1}{x^{2}}=83,$ then $x^{3}-\frac{1}{x^{3}}$ is

If $\frac{x-a^{2}}{b+c}+\frac{x-b^{2}}{c+a}+\frac{x-c^{2}}{a+b}=4(a+b+c),$ then $x$ is equal to:

$\frac{0.0203 \times 2.92}{0.0073 \times 14.5 \times 0.7} = ?$

Out of the given responses,one of the factors of $(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module