Is (lnx)^2 equivalent to ln^2 x?

Is there even such a thing as ${\ln}^{2} x$?

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Explanation

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Explanation:

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Alan N. Share
Sep 23, 2017

Yes, but also see below

Explanation:

${\ln}^{2} x$ is simply another way of writing ${\left(\ln x\right)}^{2}$ and so they are equivalent.

However, these should not be confused with $\ln {x}^{2}$ which is equal to $2 \ln x$

There is only one condition where ${\ln}^{2} x = \ln {x}^{2}$ set out below.

${\ln}^{2} x = \ln {x}^{2} \to {\left(\ln x\right)}^{2} = 2 \ln x$

$\therefore \ln x \cdot \ln x = 2 \ln x$

Since $\ln x \ne 0$

$\ln x \cdot \cancel{\ln} x = 2 \cdot \cancel{\ln} x$

$\ln x = 2$

$x = {e}^{2}$

Hence, ${\ln}^{2} x = \ln {x}^{2}$ is only true for $x = {e}^{2}$

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