Question

How do you solve `(ln x)^2=ln x^2`?

Answer 1

`x=1color(white)("XXX")orcolor(white)("XXX")x=e^2`

Explanation:

Remember: `ln(x^2)=2ln(x)`

Let `k=ln(x)`

Therefore
`color(white)("XXX")(ln(x))^2=ln(x^2)`
is equivalent to
`color(white)("XXX")k^2=2k`

`color(white)("XXX")k^2-2k=0`

`color(white)("XXX")k(k-2)=0`

#color(white)("XXX"){: (k=0,color(white)("XX")orcolor(white)("XX"),k=2), (rarr ln(x)=0,,rarr ln(x)=2), (rarr e^0=x,,rarr e^2=x), (rarr x=1,,rarr x=e^2) :}#