Question

simplify.
$8^{-\frac{6}{5}} cdot 4^{-\frac{6}{5}}$

Explanation:

Step1: Rewrite bases as powers of 2

$8=2^3$, $4=2^2$, so substitute:
$8^{-\frac{6}{5}} \cdot 4^{-\frac{6}{5}} = (2^3)^{-\frac{6}{5}} \cdot (2^2)^{-\frac{6}{5}}$

Step2: Apply exponent power rule

Use $(a^m)^n=a^{m \cdot n}$:
$(2^3)^{-\frac{6}{5}} = 2^{3 \cdot (-\frac{6}{5})}=2^{-\frac{18}{5}}$, $(2^2)^{-\frac{6}{5}}=2^{2 \cdot (-\frac{6}{5})}=2^{-\frac{12}{5}}$

Step3: Multiply with same base

Use $a^m \cdot a^n=a^{m+n}$:
$2^{-\frac{18}{5}} \cdot 2^{-\frac{12}{5}} = 2^{-\frac{18}{5} + (-\frac{12}{5})}=2^{-\frac{30}{5}}=2^{-6}$

Step4: Simplify negative exponent

Use $a^{-n}=\frac{1}{a^n}$:
$2^{-6}=\frac{1}{2^6}=\frac{1}{64}$

Answer:

$\frac{1}{64}$