# If $2^{k+1} = 2 \cdot 2^k$ what does $2^{k-1}$ equal?

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 $$2^{k-2} = \frac{2^k}{2^2}$$?

• You are correct in you reasoning – Hyperion Feb 13 at 17:38
• Yes, exactly. But the tag 'algebraic geometry' is not appropriate. – Berci Feb 13 at 17:39

## 3 Answers

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.)

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

Remember that

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

Then

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

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

Exactly, just keep in mind that

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