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My teacher gets the following: $(x^{2})^{12-k} = x^{2k-24}$

Where I get the following: $(x^{2})^{12-k} = x^{24-2k}$

I'd like to think of $2(12-k)$ as $2*12 - 2*k$ or $-2k + 24$. Why/how am I wrong?

He did the following: $x^{4k} * {a^{12-k} \over (x^{2})^{12-k}} = x^{4k} * a^{12-k} * x^{2k-24}$

Hum
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  • Note: $(a^b)^c = a^{bc}$. But without parenthesis, $a^{b^c}$ means $a^{(b^c)}$. Which one are you talking about, and what is the context? – Arturo Magidin Feb 13 '12 at 04:02
  • You and your teacher are both wrong, which probably means that you have not accurately reported what your teacher actually did. – Gerry Myerson Feb 13 '12 at 04:04
  • Sorry, clarified. – Hum Feb 13 '12 at 04:05
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    No, not clarified: corrected. If you are now accurately reporting what your teacher got, your teacher made a mistake. – Gerry Myerson Feb 13 '12 at 04:08
  • If everything is precisely as you reported (after the edit) then you are right. – André Nicolas Feb 13 '12 at 04:10
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    @GerryMyerson I think you're being a little unpleasant. I think it is quite logical to expect him to be learning $$(a^b)^c = a^{bc}$$. – Pedro Feb 13 '12 at 04:18
  • Added the example of what he did now. – Hum Feb 13 '12 at 04:20
  • @Peter, if I'm only being a little unpleasant, then I've made a lot of progress. But I'm not sure I understand your comment about $(a^b)^c=a^{bc}$. Perhaps you did not see the original version of the question, where OP had $x^{2^{12-k}}$. It is one thing to learn that $(a^b)^c=a^{bc}$, quite another to say the teacher wrote $a^{b^c}$ when he actually wrote $(a^b)^c$. – Gerry Myerson Feb 13 '12 at 05:08
  • @GerryMyerson What I mean is that it is quite unusual that expressions such as $$a^{b^{cx-d}}$$ appear in the same context as the word "teacher". – Pedro Feb 13 '12 at 05:12
  • @Peter, I'm sorry, you've lost me. I'm not convinced the combined appearance is unusual, and I don't understand how the alleged unusualness relates to my comments in this discussion. – Gerry Myerson Feb 13 '12 at 06:14
  • @GerryMyerson It's ok, it isn't a big deal anyways. – Pedro Feb 13 '12 at 12:01

1 Answers1

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Your first and last sentences contradict each other. I will assume that the last one is correct.

In that case, what the teacher is doing is $$ \frac1{(x^2)^{12-k}}=(x^2)^{k-12}=x^{2k-24} $$

Martin Argerami
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