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Powers: Reverse Rules - Raising a Power to Another Power

If $$a = 3^3 \cdot 2^9$$, $$b = 3^6 \cdot 7^3$$, and $$c = 2^6 \cdot 5^3$$, which of the following is true?
Alternative method: Since all three numbers have only third powers, take the cube root of all three and compare the cube roots. You will get the same results as above, without the cube powers. Taking the cube root out of the variables will lead to the same results without the common powers. Taking the cube root means dividing the exponents by 3. >$$\sqrt[3]{a} = \sqrt[3]{3^3 \cdot 2^9} = 3^1 \cdot 2^3 = 3 \cdot 8 = 24$$ >$$\sqrt[3]{b} = \sqrt[3]{3^6 \cdot 7^3} = 3^2 \cdot 7^1 = 9 \cdot 7 = 63$$ >$$\sqrt[3]{c} = \sqrt[3]{2^6 \cdot 5^3} = 2^2 \cdot 5^1 = 4 \cdot 5 = 20$$

Incorrect.

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Incorrect.

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Incorrect.

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Incorrect.

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Correct. [[snippet]] >$$a = 3^3 \cdot 2^9 = \left(3^1 \cdot 2^3\right)^3 = \left(3 \cdot 8\right)^3 = \left(24\right)^3$$ >$$b = 3^6 \cdot 7^3 = \left(3^2 \cdot 7^1\right)^3 = \left(9 \cdot 7\right)^3 = \left(63\right)^3$$ >$$c = 2^6 \cdot 5^3 = \left(2^2 \cdot 5^1\right)^3 = \left(4 \cdot 5\right)^3 = \left(20\right)^3$$ Now that all the variables are represented as (something)$$^3$$, they can be easily compared. Since $$63 \gt 24 \gt 20$$, then $$b \gt a \gt c$$.
$$c \gt b \gt a$$
$$b \gt a \gt c$$
$$a \gt c \gt b$$
$$b \gt c \gt a$$
$$c \gt a \gt b$$
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