Question

Need help ASAP

Answer 1

Part (1) : The solution is $729$

Part (2): The solution is $\frac{1}{16 x^{8}}$

Part (3): The solution is $\frac{2 x^{2}}{3 y z^{7}}$

Explanation:

Part (1): The expression is $3^{2} \cdot3^{4}$

Applying the exponent rule, $a^{b} \cdot a^{c}=a^{b+c}$, we get,

$3^{2} \cdot 3^{4}=3^{2+4}$

Adding the exponent, we get,

$3^{2} \cdot3^{4}=3^6=729$

Thus, the simplified value of the expression is $729$

Part (2): The expression is $\left(2 x^{2}\right)^{-4}$

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$, we have,

$\left(2 x^{2}\right)^{-4}=\frac{1}{\left(2 x^{2}\right)^{4}}$

Simplifying the expression, we have,

$\frac{1}{2^4x^8}$

Thus, we have,

$\frac{1}{16 x^{8}}$

Thus, the value of the expression is $\frac{1}{16 x^{8}}$

Part (3): The expression is $\frac{2 x^{4} y^{-4} z^{-3}}{3 x^{2} y^{-3} z^{4}}$

Applying the exponent rule, $\frac{x^{a}}{x^{b}}=x^{a-b}$, we have,

$\frac{2x^{4-2}y^{-4+3}z^{-3-4}}{3}$

Adding the powers, we get,

$\frac{2x^{2}y^{-1}z^{-7}}{3}$

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$, we have,

$\frac{2 x^{2}}{3 y z^{7}}$

Thus, the value of the expression is $\frac{2 x^{2}}{3 y z^{7}}$