# Simple Algebraic Manipulation

• May 14th 2010, 09:55 AM
1337h4x
Simple Algebraic Manipulation
Can somebody explain the rules of algebra so that the following expression makes sense:

$\displaystyle 2^{2 \cdot 2^m}+1$ $\displaystyle =\left(2^{2^m}\right)^2+1$

I thought that $\displaystyle \left(2^{2^m}\right)^2+1$ $\displaystyle =4^{ \cdot 4^(m^2)}+1$ but apparently thats wrong.
• May 14th 2010, 10:11 AM
Isomorphism
Quote:

Originally Posted by 1337h4x
Can somebody explain the rules of algebra so that the following expression makes sense:

$\displaystyle 2^{2 \cdot 2^m}+1$ $\displaystyle =\left(2^{2^m}\right)^2+1$

I thought that $\displaystyle \left(2^{2^m}\right)^2+1$ $\displaystyle =4^{ \cdot 4^(m^2)}+1$ but apparently thats wrong.

The rule is $\displaystyle (a^l)^k= a^{lk}$.

Now choose $\displaystyle a = 2, l = 2$ and $\displaystyle k = 2^m$...
• May 14th 2010, 11:22 AM
1337h4x
Quote:

Originally Posted by Isomorphism
The rule is $\displaystyle (a^l)^k= a^{lk}$.

Now choose $\displaystyle a = 2, l = 2$ and $\displaystyle k = 2^m$...

I'm not sure how we get from my form of $\displaystyle \left(a^{b^m}\right)^k$ to $\displaystyle (a^l)^k$
• May 14th 2010, 12:45 PM
masters
Quote:

Originally Posted by 1337h4x

Hi 1337h4x,

This is what it looks like to me.

$\displaystyle 2^{2 \cdot 2^m}+1=\left(2^{2^m}\right)^2$

$\displaystyle 2^{2^1 \cdot 2^m}+1=2^{2^m \cdot 2}$

$\displaystyle 2^{2^{m+1}}+1=2^{2^{m+1}}+1$

The properties you use is the power of a power rule and the product of powers rule.

$\displaystyle (a^m)^n=a^{mn}$

$\displaystyle a^m \cdot a^n=a^{m+n}$