Question

1. $\frac{11^{-7} cdot 5^{9}}{6^{-9}}$
2. $\frac{(-21)^{-4} cdot (-4)^{0}}{3^{-5} cdot 7^{-9}}$

Explanation:

Problem 8

Step1: Rewrite negative exponents

A term $a^{-n} = \frac{1}{a^n}$, so move negative exponent terms to opposite fraction part:
$\frac{11^{-7} \cdot 5^9}{6^{-9}} = \frac{5^9 \cdot 6^9}{11^7}$

Step2: Combine like exponents

Use $a^n \cdot b^n = (a \cdot b)^n$:
$\frac{(5 \cdot 6)^9}{11^7} = \frac{30^9}{11^7}$

Problem 11

Step1: Simplify zero exponent

Any non-zero term to 0 power is 1: $(-4)^0 = 1$
$\frac{(-21)^{-4} \cdot 1}{3^{-5} \cdot 7^{-9}} = \frac{(-21)^{-4}}{3^{-5} \cdot 7^{-9}}$

Step2: Rewrite negative exponents

Move negative exponent terms to opposite fraction part:
$\frac{3^5 \cdot 7^9}{(-21)^4}$

Step3: Rewrite $(-21)^4$ as product

$(-21) = (-3 \cdot 7)$, so $(-21)^4 = (3 \cdot 7)^4 = 3^4 \cdot 7^4$:
$\frac{3^5 \cdot 7^9}{3^4 \cdot 7^4}$

Step4: Subtract exponents for like bases

Use $\frac{a^m}{a^n} = a^{m-n}$:
$3^{5-4} \cdot 7^{9-4} = 3^1 \cdot 7^5$

Step5: Calculate the simplified form

$3 \cdot 16807 = 50421$

Answer:

Problem 8: $\boldsymbol{\frac{30^9}{11^7}}$
Problem 11: $\boldsymbol{50421}$ (or $\boldsymbol{3 \cdot 7^5}$)