Apply the Properties of Integer Exponents to generate equivalent expressions to (7^−3)^7⋅ 7^20/6^−2 with only positive exponents, then solve the expression. Provide the answer as an improper fraction.(1 point)
To apply the properties of integer exponents, we can first simplify the expression $(7^{-3})^7$ by multiplying the exponents: $(7^{-3})^7 = 7^{-3 \cdot 7} = 7^{-21}$.
Next, we can simplify the expression $7^{20}$ by multiplying the exponents: $7^{20} = 7^{\frac{2}{6} \cdot (-2)} = (7^{1/3})^{-2} = \left(\frac{1}{7^{-1/3}}\right)^2 = \left(\frac{1}{\sqrt[3]{7}}\right)^2 = \left(\frac{1}{\sqrt[3]{7}}\right)^{2\cdot3} = \frac{1}{7^2} = \frac{1}{49}$.
Therefore, the expression $(7^{-3})^7 \cdot 7^{20} \div 6^{-2}$ is equivalent to $7^{-21} \cdot \frac{1}{49} \div 6^{-2}$.
Now, let's simplify the expression using the quotient and power rules of exponents:
$7^{-21} \cdot \frac{1}{49} \div 6^{-2} = \frac{1}{7^{21}} \cdot \frac{1}{49} \div \frac{1}{6^2} = \frac{1}{7^{21}} \cdot \frac{1}{49} \cdot 6^2 = \frac{1}{7^{21}} \cdot \frac{1}{49} \cdot 36 = \frac{1 \cdot 1 \cdot 36}{7^{21} \cdot 49} = \frac{36}{7^{21} \cdot 49}$.
So the answer is $\frac{36}{7^{21} \cdot 49}$.
what?????
I apologize for any confusion. Let's start over.
The expression we are given is $(7^{-3})^7 \cdot 7^{20} \div 6^{-2}$.
Using the property $(a^m)^n = a^{m \cdot n}$, we can simplify $(7^{-3})^7$ as $7^{-3 \cdot 7} = 7^{-21}$.
Using the property $a^m \cdot a^n = a^{m + n}$, we can simplify the expression as $7^{-21} \cdot 7^{20} \div 6^{-2}$.
Applying the property $a^m \div a^n = a^{m - n}$, we can simplify further as $7^{-21} \cdot 7^{20} \cdot 6^2$.
Applying the property $a^{-n} = \frac{1}{a^n}$, we have $\frac{1}{7^{21}} \cdot 7^{20} \cdot 6^2$.
Simplifying $7^{20}$ as $7^{20} = 7^{\frac{3}{3} \cdot 20} = (7^{\frac{1}{3}})^{60} = (7^{1/3})^{60} = (7^{1/3})^{2 \cdot 30} = (7^{1/3})^{2 \cdot 3 \cdot 10} = (7^{1/3})^{6 \cdot 10} = (7^{1/3})^{60} = (\sqrt[3]{7})^{60} = (\sqrt[3]{7})^{3 \cdot 20} = (\sqrt[3]{7})^{20} = \left(\frac{1}{\sqrt[3]{7^{-1}}}\right)^{20} = \left(\frac{1}{\sqrt[3]{7}}\right)^{20} = \left(\frac{1}{\sqrt[3]{7}}\right)^{3 \cdot 6 + 2} = \left(\frac{1}{\sqrt[3]{7}}\right)^{3 \cdot 6} \cdot \left(\frac{1}{\sqrt[3]{7}}\right)^2 = (7^{\frac{1}{3}})^{-6} \cdot (7^{\frac{1}{3}})^{-2} = 7^{-6} \cdot 7^{-2} = \frac{1}{7^6} \cdot \frac{1}{7^2} = \frac{1}{7^6 \cdot 7^2} = \frac{1}{7^8}$.
Therefore, the expression simplifies to $\frac{1}{7^{21}} \cdot \frac{1}{7^8} \cdot 6^2 = \frac{1}{7^{21}} \cdot \frac{1}{7^8} \cdot 36 = \frac{36}{7^{21} \cdot 7^8} = \frac{36}{7^{21+8}} = \frac{36}{7^{29}}$.
So the answer is $\frac{36}{7^{29}}$.
I apologize for any confusion caused by my previous response.
ANSWER MORE CLAR
I apologize for any confusion. Let's go through the steps again in a clearer manner.
The expression we are given is $(7^{-3})^7 \cdot 7^{20} \div 6^{-2}$.
First, let's simplify $(7^{-3})^7$. Using the property $(a^m)^n = a^{m \cdot n}$, we have $(7^{-3})^7 = 7^{-3 \cdot 7} = 7^{-21}$.
Next, let's simplify $7^{20}$.
To do this, we can rewrite $7^{20}$ as $7^{6 \cdot \frac{20}{6}}$, where $\frac{20}{6}$ is the same as $\frac{2}{6} \cdot (-2)$.
Using the property $a^{m/n} = \sqrt[n]{a^m}$, we simplify $\frac{2}{6}$ as $\sqrt[3]{7^2}$.
Therefore, $7^{20} = 7^{6 \cdot \sqrt[3]{7^2}} = (7^6)^{\sqrt[3]{7^2}} = (\sqrt[3]{7^6})^2 = 7^2 = 49$.
Now, let's simplify $6^{-2}$. Using the property $a^{-n} = \frac{1}{a^n}$, we have $6^{-2} = \frac{1}{6^2} = \frac{1}{36}$.
Putting it all together, our expression becomes $7^{-21} \cdot 49 \div \frac{1}{36}$.
Using the property $a^{-n} = \frac{1}{a^n}$, we rewrite $7^{-21}$ as $\frac{1}{7^{21}}$.
The expression now becomes $\frac{1}{7^{21}} \cdot 49 \div \frac{1}{36}$.
Finally, we simplify the expression by dividing 49 by 1 and multiplying by 36:
$\frac{1}{7^{21}} \cdot 49 \div \frac{1}{36} = \frac{49}{1} \cdot \frac{36}{7^{21}} = \frac{49 \cdot 36}{7^{21}} = \frac{1764}{7^{21}}$.
So the answer is $\frac{1764}{7^{21}}$.
To generate equivalent expressions with positive exponents, we will use the property of exponentiation that states (a^m)^n = a^(m*n).
Step 1: Simplify the expression (7^(-3))^7.
(7^(-3))^7 = 7^(-3 * 7) = 7^(-21)
Step 2: Simplify the expression 7^20.
7^20 = 7^(20 * 1) = 7^20
Step 3: Simplify the expression 6^(-2).
6^(-2) = 1/(6^2) = 1/36
Now we can rewrite the given expression using positive exponents:
(7^(-3))^7⋅ 7^20/6^(-2) = 7^(-21) ⋅ 7^20 ⋅ (1/36)
To simplify, we add the exponents when multiplying and subtract the exponents when dividing powers of the same base:
7^(-21) ⋅ 7^20 ⋅ (1/36) = 7^(-21 + 20) ⋅ (1/36) = 7^(-1) ⋅ (1/36) = 1/7 ⋅ (1/36) = 1/(7⋅36) = 1/252
Therefore, the solution is 1/252 as an improper fraction.
To generate equivalent expressions with only positive exponents, we can apply the properties of integer exponents, specifically:
1. For any nonzero number a, a^(-n) = 1/a^n - This property allows us to rewrite negative exponents as positive exponents by reciprocating the base.
Now let's break down the expression step-by-step.
Step 1: Simplify (7^(-3))^7.
Using the property stated above, we can convert the negative exponent to a positive exponent:
(7^(-3))^7 = (1/7^3)^7 = 1/7^(3*7) = 1/7^21
Step 2: Simplify 7^20/6^(-2).
Again, we need to convert the negative exponent to a positive exponent:
6^(-2) = 1/6^2 = 1/36
Now we can simplify the overall expression:
(7^(-3))^7 * 7^20/6^(-2) = (1/7^21) * (7^20 / 1/36)
Next, we multiply the two fractions together:
(1/7^21) * (7^20 / 1/36) = (1 * 7^20) / (7^21 * 1/36)
Simplifying further:
= (7^20) / (7^21 * 1/36)
Now let's simplify the exponents:
= (7^(20 - 21)) / (1/36)
= (7^(-1)) / (1/36)
To change the negative exponent to a positive exponent, we use the property stated above:
= 1/7^1 * 36
= 1/7 * 36
Finally, we simplify:
= 36/7
Therefore, the answer as an improper fraction is 36/7.