Is ${(ln x)}^{2}$ $(lnx)^2$ equivalent to ${ln}^{2} x$ $ln^2 x$ ?

Question

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

233099 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-23T04:20:19

Yes, but also see below

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

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

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

${ln}^{2} x = {ln x}^{2} \to {(ln x)}^{2} = 2 ln x$ $ln^2 x = ln x^2 -> (lnx)^2 = 2lnx$

$:. lnx * lnx = 2lnx$

Since $lnx !=0$

$lnx * cancel lnx = 2 * cancel lnx$

$lnx = 2$

$x =e^2$

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