# Exponential Problem

• Dec 3rd 2012, 11:12 PM
BobBali
Exponential Problem
Hi,
Good day All,

The problem asks to simplify without using a calculator:

$\displaystyle \frac{2^{14} - 2^{11} + 7}{2^{11} + 7}$

$\displaystyle \frac{2^{11} (2^{3} - 1) + 7}{2^{11} + 1}$

$\displaystyle \frac{2^{11} (8-1) + 7}{2^5 2^5 2^1}$

$\displaystyle \frac{2^5 2^5 2^1 (7) +7}{2^5 2^5 2^1 + 1}$

Not sure if i'm on the right track....?
• Dec 3rd 2012, 11:25 PM
MarkFL
Re: Exponential Problem
I assume you mean:

$\displaystyle \frac{2^{14}-2^{11}+7}{2^{11}+1}=\frac{2^{11}(2^3-1)+7}{2^{11}+1}=\frac{7(2^{11}+1)}{2^{11}+1}=7$
• Dec 4th 2012, 01:12 AM
puresoul
Re: Exponential Problem
Quote:

Originally Posted by MarkFL2
I assume you mean:

$\displaystyle \frac{2^{14}-2^{11}+7}{2^{11}+1}=\frac{2^{11}(2^3-1)+7}{2^{11}+1}=\frac{7(2^{11}+1)}{2^{11}+1}=7$

sorry, I am not the one who posted this question but I am just curious...
How did you get 7(2^11 +1)/ 2^11 +1 from 2^11(2^3 -1) +7 / 2^11 +1 ???
• Dec 4th 2012, 01:23 AM
MarkFL
Re: Exponential Problem
I will add an intermediary step:

$\displaystyle \frac{2^{14}-2^{11}+7}{2^{11}+1}=\frac{2^{11}(2^3-1)+7}{2^{11}+1}=\frac{2^{11}\cdot7+7}{2^{11}+1}= \frac{7(2^{11}+1)}{2^{11}+1}=7$
• Dec 4th 2012, 01:25 AM
puresoul
Re: Exponential Problem
Quote:

Originally Posted by MarkFL2
I will add an intermediary step:

$\displaystyle \frac{2^{14}-2^{11}+7}{2^{11}+1}=\frac{2^{11}(2^3-1)+7}{2^{11}+1}=\frac{2^{11}\cdot7+7}{2^{11}+1}= \frac{7(2^{11}+1)}{2^{11}+1}=7$

Oh..
I didn't realise that 2^3 -1 is 7
Thnks ^^