Question

simplify.
$4^{-\frac{2}{3}} cdot 16^{-\frac{2}{3}}$

Explanation:

Step1: Rewrite bases as powers of 2

$4=2^2$, $16=2^4$
So $4^{-\frac{2}{3}}=(2^2)^{-\frac{2}{3}}$, $16^{-\frac{2}{3}}=(2^4)^{-\frac{2}{3}}$

Step2: Simplify exponents via power rule

Power rule: $(a^m)^n=a^{m \cdot n}$
$(2^2)^{-\frac{2}{3}}=2^{2 \cdot (-\frac{2}{3})}=2^{-\frac{4}{3}}$
$(2^4)^{-\frac{2}{3}}=2^{4 \cdot (-\frac{2}{3})}=2^{-\frac{8}{3}}$

Step3: Multiply terms, add exponents

Product rule: $a^m \cdot a^n=a^{m+n}$
$2^{-\frac{4}{3}} \cdot 2^{-\frac{8}{3}}=2^{-\frac{4}{3} + (-\frac{8}{3})}=2^{-\frac{12}{3}}=2^{-4}$

Step4: Rewrite negative exponent as reciprocal

$2^{-4}=\frac{1}{2^4}$

Step5: Compute the final value

$2^4=16$, so $\frac{1}{2^4}=\frac{1}{16}$

Answer:

$\frac{1}{16}$