Apply the properties of integer exponents to generate equivalent expressions to (7^-3)^7X7^20/6^-2 with positive exponents then solve the expression

Answer 1

To simplify the expression $(7^{-3}{)}^7\times 7^{20}/6^{-2}$ and express it with positive exponents, we'll use the properties of integer exponents.

1. Apply the power of a power rule: $(a^m{)}^n=a^{m\cdot n}$. $$(7^{-3}{)}^7=7^{-3\cdot 7}=7^{-21}$$

2. Combine the terms with the same base: We now have $7^{-21}\times 7^{20}$. Using the property $a^m\times a^n=a^{m+n}$: $$7^{-21}\times 7^{20}=7^{-21+20}=7^{-1}$$

3. Convert the exponent to positive: Remember that $a^{-n}=\frac{1}{a^n}$: $$7^{-1}=\frac{1}{7}$$

4. Simplify the denominator: Now we simplify $6^{-2}$ as well: $$6^{-2}=\frac{1}{6^2}=\frac{1}{36}$$

5. Combine everything into a single expression: $$\frac{1}{7}÷\frac{1}{36}$$

6. Dividing by a fraction is the same as multiplying by its reciprocal: $$\frac{1}{7}\times 36=\frac{36}{7}$$

Thus, the expression $(7^{-3}{)}^7\times 7^{20}/6^{-2}$ simplifies and evaluates to: $$\frac{36}{7}$$