If $$6^x > 216^2$$, what is the least possible integer value of $$x$$?

Correct.
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Since $$216 = 6 \cdot 36 = 6 \cdot 6 \cdot 6$$, you can express $$6^x > 216^2$$ as
>$$6^x > (6^3)^2$$
>$$6^x > 6^6$$.
Since the bases are the same, ignore the bases and compare the exponents.
>$$x > 6$$
Hence, the least possible integer value of $$x$$ is 7.

Incorrect.
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Incorrect.
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Incorrect.
Note that the question uses an inequality, not an equation.
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Incorrect.
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9

7

6

5

3