# Simplifying an expression with 2's raised to powers

• Mar 18th 2009, 04:39 AM
_icebox
Simplifying an expression with 2's raised to powers
Hey people!

$\displaystyle \frac{2^n+^4 - 2.2^n} {2.2^n+^3}$

It's easy?!

I just can't transform those stuffs in numbers (Headbang) !!!
• Mar 18th 2009, 05:13 AM
Quote:

Originally Posted by _icebox
Hey people!

$\displaystyle \frac{2^n+^4 - 2.2^n} {2.2^n+^3}$

It's easy?!

I just can't transform those stuffs in numbers (Headbang) !!!

$\displaystyle \frac{2^{n+4} - 2\times2^n} {2.2^{n+3}}$

$\displaystyle a^{x} \times a^{y} = a^{x+y}$

$\displaystyle =\frac{2^{n+4} - 2^{n+1}} {2^{n+4}}$

$\displaystyle =\frac{2^{n+4} - 2^{n+1}} {2^{n+4}}$

Divide numerator and denominator by $\displaystyle 2^{n+1}$

$\displaystyle =\frac{2^{3} - 1} {2^{3}}$

$\displaystyle =\frac{8 - 1} {8}$

$\displaystyle =\frac{7} {8}$
• Mar 18th 2009, 05:22 AM
masters
Quote:

$\displaystyle \frac{2^{n+4} - 2\times2^n} {2.2^{n+3}}$

$\displaystyle a^{x} \times a^{y} = a^{x+y}$

$\displaystyle =\frac{2^{n+4} - 2^{n+1}} {2^{n+4}}$

$\displaystyle =\frac{2^{n+4} - 2^{n+1}} {2^{n+4}}$

Divide numerator and denominator by $\displaystyle 2^{n+1}$

$\displaystyle =\frac{2^{3} - 1} {2^{3}}$

$\displaystyle =\frac{8 - 1} {8}$

$\displaystyle =\frac{7} {8}$

I'm impressed that you could interpret $\displaystyle 2.2^n$ as $\displaystyle 2 \cdot 2^n$.

I was having a tough time fooling around with the decimal.
• Mar 18th 2009, 05:25 AM
Quote:

Originally Posted by masters
I'm impressed that you could interpret $\displaystyle 2.2^n$ as $\displaystyle 2 \cdot 2^n$.

I was having a tough time fooling around with the decimal.

The time I saw that + sign jumping up and down , I knew interpretation is the keyword :p ..(Giggle)
• Mar 18th 2009, 05:27 AM
_icebox
Yes! That's it!!! (Clapping)

I've some problems with the Latex... I don't know exacly how to use that hauhauhaha

But that's exacly what I was looking for =D
I stopped when it was necessary to divide numerator and denominator =P