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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

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

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    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 sides2011-10-17

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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.

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    Oh ... icic, I looked into the textbook errata and discovered its indeed an error ... caused me so much confusion :)2011-10-17