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I'm stuck in the follow equation: $$\dfrac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}$$

As all the bases are equal, I got $\dfrac{3n + 5}{2n - 3}$

Where I've to go now ?

Thanks

EDIT:

Then, my initial idea was totally wrong, starting again, in the right way I got:

$$\dfrac{2^n(2^4 + 2^2 + 2^{-1})}{2^n(2^{-2} + 2^{-1})}$$

but it's still wrong, I didn't get the right idea on the divisions you have shown to me in the answers.

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1  
It seems like you added the exponents in the numerator and the denominator, which you cannot do because you don't have a product: $b^n \times b^m = b^{n+m} \neq b^n + b^m$, and you got rid of the bases. –  talmid Jun 22 '12 at 21:20
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@aajjbb: What you added is perfectly correct. Now cancel the $2^n$. But in order to get something that "looks nice" (integer divided by integer) it is helpful to multiply your new top and bottom by $4$ (that is, $2^2$). –  André Nicolas Jun 22 '12 at 23:09

2 Answers 2

up vote 4 down vote accepted

You don't have an equation, you are probably asked to simplify. Divide top and bottom by $2^{n-2}$.

When you do that, at the bottom you will have $1+2$. On top you will have $2^6+2^4+2$.

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I've understood all the logic behind the solutions, but I'm not good yet with this concept of division between $\dfrac{2^{n + 4}}{2^{n - 2}}$ could you explain the steps to me ? Thanks –  aajjbb Jun 22 '12 at 22:23
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There are several ways to think of it. You can say that by the rules of exponentiation, $2^{n+4}=2^{n-2}2^6$. Now we can cancel $2^{n-2}$ on top and bottom. Or else we can view $2^{n+4}$ as the product of $n+4$ $2$'s, and $2^{n-2}$ as the product of $n-2$ $2$'s, which is $6$ fewer. So when we divide we get the product of $6$ $2$'s left, that is, $2^6$. However, I think that the "factoring" explanation of Brian M. Scott may be more directly suited to your needs. –  André Nicolas Jun 22 '12 at 22:54

You’ve nothing to solve. My guess is that you’re supposed to simplify the fraction:

$$\begin{align*} \dfrac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}&=\frac{2^{n-1}(2^5+2^3+1)}{2^{n-2}(1+2)}\\ &=\frac{2^{n-1}}{2^{n-2}}\cdot\frac{32+8+1}{3}\;, \end{align*}$$

and you should have no trouble finishing it.

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Typo in the denominator. –  talmid Jun 22 '12 at 21:21
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@talmid: Thanks. (At least I was consistent after the typo!) –  Brian M. Scott Jun 22 '12 at 21:22

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