Exponentiation involving high powers of 2.

Question

How to calculate

$2^{987}-2^{986}=?$

without using a calculator?

Answer: $2^{986}$

Answer 1

It appears that you mean to ask "$2^{987}-2^{986}=?$". Note that $2^{987}=2\cdot 2^{986}$ and that $2x-x = x$.

So:

$$2^{987}-2^{986} = 2\cdot 2^{986}-2^{986} = (2-1)2^{986} = 1\cdot 2^{986}=2^{986}$$

Answer 2

Notice, the following steps $$2^{987}-2^{986}$$$$\iff 2^{986+1}-2^{986} $$$$\iff 2\times 2^{986}-2^{986}$$$$\iff 2^{986}(2-1)=\color{red}{2^{986}}$$

Answer 3

In greater generality, $b^n-b^m$ (for $n>m$) equals $$b^{(n-m)+m}-b^m$$ $$=b^{n-m} \times b^m-b^m$$ $$=(b^{n-m}-1) \times b^m$$ as, for example, $10^9-10^6=999\times 10^6$.  If $n-m=1$, this clearly reduces to $$(b-1) \times b^m$$ and, if (additionally) $b=2$, it reduces further to $$b^m\ \ \ \ \ \ \ \ i.e.,\ \ 2^m$$