If $$\displaystyle 3^{-5x} < \frac{1}{243^2}$$, what is the smallest possible integer value of $$x$$?

Correct.

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First, convert the negative power of $$3$$ into a positive one (i.e., $$3^{-5x} = \frac{1}{3^{5x}}$$). >$$\displaystyle \frac{1}{3^{5x}} < \frac{1}{243^2}$$ Cross multiply to get $$243^2 < 3^{5x}$$. Express everything as a power of $$3$$. >$$243 = 9 \cdot 27 = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^5$$ So $$243^2 < 3^{5x}$$ can be expressed as >$$(3^{5})^{2} < 3^{5x}$$ >$$3^{10} < 3^{5x}$$. Since the bases are the same, the powers can be compared directly. >$$10 < 5x$$ >$$2 < x$$ Hence, the smallest possible integer value of $$x$$ is $$3$$.Incorrect.

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It's true that $$x=6$$ will indeed make the left side smaller than the right side, but did you choose this answer because you missed out on the fact that the question asks for theIncorrect.

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Incorrect.

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Incorrect.

Note that the question uses an inequality, not an equation. If you've done your calculation right, $$x$$ cannot equal $$2$$.

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Read the question carefully, and form a strategy before you dive in. The question calls for the smallest value of $$x$$. It actually pays to start plugging in from the smallest value, and moving up until you find the first answer choice that makes the left side of the inequality smaller than the right side.

Care to leave a comment and tell us about it?

$$2$$

$$3$$

$$4$$

$$5$$

$$6$$

Yes, I got lost in trying to make the left side smaller than the right side.

No, I made some other error.