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Its been too long since my highschool algebra class, and now I need those skills again. I'm reviewing all my math on khanacademy.org but I need this answered sooner than that.

What algebraic property(s) makes the transformation between the following two equations? It would be really helpful if someone wrote it out in excruciatingly small steps.

$$\begin{align*} V = {} & \left(1+f2^{-n}\right)2^E \\ V = {} & \left(2^n+f\right)2^{E-n} \end{align*}$$

Bart Michels
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TheEntity08
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    Insert $2^n\cdot 2^{-n}$ after the closing parentheses and apply the distributive property. – egreg Oct 03 '13 at 19:25

2 Answers2

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$$V=(1+f*2^{-n})*2^E$$

We can rewrite this as:

$V=(1+\dfrac{f}{2^n})*2^E$

Finding a common denominator, we have:

$V=\left(\dfrac{2^n + f}{2^n}\right)*2^E$

We can write this as:

$V=(2^n + f)*\dfrac{2^E}{2^n} = (2^n + f)*2^{E-n}$

Amzoti
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$$ (1+f2^{-n})2^E = \Big((1+f2^{-n})\cdot2^n\Big)\Big(2^{-n}2^E\Big) $$ This is true because $2^n\cdot2^{-n}=1$; they cancel each other.

Michael Hardy
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