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If $$p = 2^8 \cdot3^4$$, $$q = 5^2 \cdot2^4$$, and $$r = 3^4 \cdot 5^2$$, which of the following must be true?
Here's an alternative method: Since all three numbers have only even powers, take the square root of all three and compare the square roots. You will get the same results as above, without the square powers. Taking the square root out of the variables will lead to the same results without the common powers. Taking the square root means dividing the exponents by 2. >$$\sqrt{p} = \sqrt{2^8 \cdot 3^4} = 2^4 \cdot 3^2 = 16 \cdot 9 = 144$$ >$$\sqrt{q} = \sqrt{5^2 \cdot 2^4} = 5^1 \cdot 2^2 = 5 \cdot 4 = 20$$ >$$\sqrt{r} = \sqrt{3^4 \cdot 5^2} = 3^2 \cdot 5^1 = 9 \cdot 5 = 45$$
Correct. [[snippet]] Rewrite each variable as a number squared. >$$p = 2^8 \cdot 3^4 = \left(2^4 \cdot 3^2\right)^2 = \left(16 \cdot 9\right)^2 = \left(144\right)^2$$ >$$q = 5^2 \cdot 2^4 = \left(5^1 \cdot 2^2\right)^2 = \left(5 \cdot 4\right)^2 = \left(20\right)^2$$ >$$r = 3^4 \cdot 5^2 = \left(3^2 \cdot 5\right)^2 = \left(9 \cdot 5\right)^2 = \left(45\right)^2$$ Now that all the variables are represented as (something)$$^2$$, they can be easily compared. Since $$144 \gt 45 \gt 20$$, then $$p \gt r \gt q$$.
$$p \gt r \gt q$$
$$q \gt p \gt r$$
$$q \gt r \gt p$$
$$r \gt p \gt q$$
$$p \gt q \gt r$$