I have $a^{n-k}, \; a=x^2$. When I substitute, do I get $x^{2(n-k)}$ or $x^{2^{n-k}}$? I'm confused about which one is correct.
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The correct substitution goes as follows:
$$a^{n-k}=(x^2)^{n-k}=x^{2(n-k)}$$
The key here is that $(a^b)^c=a^{bc}$. Notably, exponents aren't associative, so $a^{(b^c)}\neq (a^b)^c$.
Stella Biderman
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If $n=k$, then
$$a^{n-k}=a^0=1$$
This is regardless of what $a$ is. Thus, it shouldn't matter what happens when we let $a=x^2$:
$$x^{2(n-k)}=x^{2(0)}=x^0=1\quad\color{green}{\checkmark\text{right}}$$
$$x^{2^{n-k}}=x^{2^0}=x^1\ne1\implies\color{red}{\text{wrong}}$$
Stella Biderman
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Simply Beautiful Art
- 74,685
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This is a good way to test two potential answers, but it's important to keep in mind that it can only rule out potential answers. It doesn't demonstrate that the remaining answer is correct. $x^{10(n-k)^2}$ satisfies this criterion too. – Stella Biderman Mar 16 '17 at 02:41
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:-) that is true, but it is the simplest way to rule out possible answers. – Simply Beautiful Art Mar 16 '17 at 12:24