# 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:

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

I thought that $\left(2^{2^m}\right)^2+1$ $=4^{ \cdot 4^(m^2)}+1$ 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:

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

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

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

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

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

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

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

Originally Posted by 1337h4x

Hi 1337h4x,

This is what it looks like to me.

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

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

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

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

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

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