How does $2^n + 2^n = 2^{n+1}$?

Question

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?

Answer 1

$2^n + 2^n = (1+1)*2^n = (2^1)*(2^n) = 2^{(1+n)}$

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 $n$.

Answer 2

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

Answer 3

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 $$3^n+3^n+3^n=3^{n+1}$$

Answer 4

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, $$\underbrace{m^n + m^n + \ldots + m^n}_{m \text{ terms}} = m(m^n) = m^{n+1}.$$

Answer 5

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, $$2^n + 2^n = 2\times 2^n = 2^{n+1}$$

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

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