In my textbook, there's something like:
$\ln(\sin 2x)^{\ln x} = \ln x \cdot \sin 2x$
I thought it should be
$\ln x\cdot \ln(\sin 2x).$
In my textbook, there's something like:
$\ln(\sin 2x)^{\ln x} = \ln x \cdot \sin 2x$
I thought it should be
$\ln x\cdot \ln(\sin 2x).$
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.