How does $2^n + 2^n = 2^{n+1}$? What property of exponents can be used to show that 
$$2^n + 2^n = 2^{n+1}$$
Does this work for all constants raised to a variable exponent?
 A: $$2^n+2^n=2 \cdot 2^n=2^{n+1}$$
A: $2^n + 2^n = (1+1)*2^n = (2^1)*(2^n) = 2^{(1+n)}$
The properties used here were $a^1=a$ and $a^{(b+c)} = (a^b)*(a^c)$.
$a^n+a^n = a^{(n+1)}$ only if $a = 2$ (or $0$, if $n>0$), but $a*a^n = a^{(n+1)}$ for all $a$ and $n$.
A: The simple answer to the second part of the question is no. For example, 
$3^n + 3^n \ne 3^{n+1}.$
For any positive integer $m$, however, 
$$\underbrace{m^n + m^n + \ldots + m^n}_{m \text{ terms}} = m(m^n) = m^{n+1}.$$
A: You have $$2^n+2^n=2^n(1+1)=2^n \cdot 2=2^{n+1}$$  It will work for other constants as long as there are as many terms as the constant.  So $$3^n+3^n+3^n=3^{n+1}$$
A: I still remember trying to make my friends understand how we can take $2^n$ outside, when I was in middle school.
$$2^n + 2^n \\
= 2(2^{n-1} + 2^{n-1}) \\
= 2^2(2^{n-2} + 2^{n-2}) \\
= 2^3(2^{n-3} + 2^{n-3})\\
\dots\\ \dots
$$
$$=2^n(2^{n-n} + 2^{n-n})\\
= 2^n(2^0 + 2^0)\\
= 2^n(1 + 1)\\
= 2^n\cdot 2^1\\
= 2^{n+1}$$
But really, all they had to understand was that multiplication is repeated addition (something which they knew but didn't know how to apply):
$$a \times b =  \sum^a b = \underbrace{b + b + b + \dots}_{a \text{ times}} $$
So, naturally,
$$2^n + 2^n = 2\times 2^n = 2^{n+1}$$

Now, there are an infinite number of similar equations to this one you find interesting:
$$ 3^n + 3^n + 3^n = 3^{n +1}\\
4^n + 4^n + 4^n + 4^n = 4^{n+1}\\
5^n + 5^n + 5^n + 5^n + 5^n = 5^{n+1}\\
\dots \\ \dots$$

Generally,
$$ \sum^a a^n = a^{n+1} \quad \forall\space n>0$$

Please note, $a^n + a^n = a^{n+1} $ is only true if $a = 0 ,2$ and $n>0$
$$[\because 0^n + 0^n = 0 + 0 = 0 = 0^{ \text{potatoes} } = 0^{ \text{tomatoes} } = 0^{n+1}]$$
