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I know that $2^{k+1} = 2 \cdot 2^k$, but what does $2^{k-1}$ equal? Is it $\frac{2^k}{2}$? Then does $2^{k-2} = \frac{2^k}{2^2}$?

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    $\begingroup$ You are correct in you reasoning $\endgroup$ – Hyperion Feb 13 at 17:38
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    $\begingroup$ Yes, exactly. But the tag 'algebraic geometry' is not appropriate. $\endgroup$ – Berci Feb 13 at 17:39
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Yes, you are thinking in the correct direction. Also this is true (a, b, c, m are taken real numbers and a is not equal to zero.)

$$a^{b.c-m}=\frac{a^{b.c}}{a^m}$$

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Remember that

$$a^{-b}=\frac1{a^b}.$$

Then

$$a^{b+c}=a^ba^c$$ generalizes to

$$a^{b-c}=a^ba^{-c}=\frac{a^b}{a^c}.$$

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Exactly, just keep in mind that

$$x^{\frac{a-b}{c}}=\sqrt[\leftroot{-1}\uproot{2}\scriptstyle c]{\frac{x^a}{x^b}}$$

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