logarithm and square rooting?

you seem to have got the answering procedure from many others . I'll just explain why you are getting both the signs . the reason is that
Sq rt(x^2) is given by Modulus of x ( mod(x) ) . Thats why you're getting both postive and negative answers due to the modulus as
mod(2) = 2 = mod(-2) . hope you got it.

i might have helped you in case you asked a thank you to remedy those and not purely the respond, besides the indisputable fact that this is obtrusive which you do no longer even want to be taught a thank you to handle that form of math issues so i do no longer think of every person's gonna grant help to.

Re-write the expression using the exponential form :

e^9 = x²

So x = ±√(e^9) = ± (e^9/2)

(√n is equivalent to n^½ )

First, note that "ln" is logarithm in base "e". So, using the properties of logarithms that: (2)log(x) = 3 equals to 2^3=x, we have:
ln(x^2) = 9
e^9 = x^2
As the solution for x^2 = 4 is +sqrt(4) and -sqrt(4), we also get for this case, the value of x is:
x = +sqrt(e^9) = +e^(9/2)
or
x = -sqrt(e^9) = -e^(9/2)
Hope that helps.

e^(lnx^2) = e^9
x^2 = e^9
x = (e^9)^(1/2) or - (e^9)^(1/2)

ln(x^2) = 9
x^2 = e^9
x = ±√(e^9)
x = ±√[(e^4)^2 * e]
x = ±e^4√e

ln(x^2) = 9
x^2 = e^9
x = +/- sqrt(e^9) = +/- e^(9/2)