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In my textbook, there's something like:

$$\ln(\sin 2x)^{\ln x} = \ln x \cdot \sin 2x$$

I thought it should be

$$\ln x\cdot \ln(\sin 2x).$$

yunone
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Jiew Meng
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    $\ln(a^b)=b\ln a$. – André Nicolas Oct 17 '11 at 07:06
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    You're absolutely correct, $\ln\left((\sin2x)^{\ln x}\right)=(\ln x)\cdot \ln (\sin 2x)$. Do you have more context? – anon Oct 17 '11 at 07:10
  • Hmm ... but according to the book, notice the 2nd $ln$ is missing. $ln(sin2x)$ is just $sin2x$. As for more context, this is actually to differenciate $sin2x^{lnx}$. This is the 1st step when $ln$ is taken on both sides – Jiew Meng Oct 17 '11 at 07:17

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If the textbook means to imply that $$\ln\left[(\sin 2x)^{\ln x}\right]=(\ln x) \cdot \sin 2x,\quad\;\;\text{ }$$ then it is in error, because in actuality it is as you say it is: $$\ln\left[(\sin 2x)^{\ln x}\right]=(\ln x)\cdot \ln(\sin2x).$$ If the rest of the derivation proceeds from this folly, it is completely wrong.

anon
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  • So the crucial question is, what's the rest of the derivation? If the argument proceeds as if the author(s) had written the right thing, then it's just a typo. If the argument seriously makes use of $\log x\sin2x$, then, as anon suggests, it's hard to imagine anything but nonsense resulting. So: what does the book do next? For that matter, what book is it? what page? – Gerry Myerson Oct 17 '11 at 11:44
  • Oh ... icic, I looked into the textbook errata and discovered its indeed an error ... caused me so much confusion :) – Jiew Meng Oct 17 '11 at 12:05