Little Logs confusion. (1 Viewer)

[RishBonjour]

had my 2u trial today.
there was a questions like:

2ln(3x^(1/2)) =...

I got my final answers as something, and a -3

Now, the -3 satisfied the RHS

but for 2ln(3x^(1/2))

It only satisfies the equation if 2 is taken to be the power in side i.e. ln((3x^(1/2))^2)

so is -3 a solution?

 

[Alkanes]

What did the question actually ask?

 

[RishBonjour]

Just realised I wrote the question wrong it was simply a solve for x question. (can't remember it exactly)

2ln(xroot3)=ln(something)

basic quadratics, but I got a bit stuffed about the negative answer :/ It was only valid if I raised everything in ln to the power of 2

 

[Shadowdude]

?

 

[D94]

No, -3 is not a solution.

 

[iBibah]

No, -3 is not a solution.

This, because it does not satisfy the original equation given.

 

[RishBonjour]

dammnnnnnnnn

but, why can't the 2 be raised inside?

then it would be ln(3x^2) and the equation becomes true for all x?

edit:
FUCK