ln(x^2) vs 2lnx ? (1 Viewer)

[kpq_sniper017]

Suppose you have to solve:
ln(2x+15)=2lnx
then the only solution is x=5 since x>0

but suppose the equation was instead:
ln(2x+15)=ln(x²)
would there then be two solutions i.e. x=5 and x=-3?

 

[Xayma]

I think there will be the two solutions.

Although Im not 100% sure I think it would work the same way as how if x²=4, x= (+-) 2.

Dont quote me but thats how I think it works.

 

[withoutaface]

To answer the question more directly:

2lnx Is exactly equal to ln(x²) What works for one works for the other.

That's why I love maths. No contradictions. If there is a contradiction, then you're using a different axiom to me.

Not necessarily, x can be negative for ln(x²), whereas for 2lnx it cannot.

So essentially 2lnx=lnx² .....(x>0)

 

[Iota]

Good point.

 

[Wild Dan Hibiki]

yeah that is a good point...

 

[withoutaface]

yeah that is a good point...

Because ln(x) is undefined for x<=0, and as such, you must keep the limitations of the first equation in your final answer.

 

[Wild Dan Hibiki]

yeah i know that but how the hell do u think of this stuff off the top of ur head?

 

[Iota]

And, taking it into consideration, I must revise my statement that x=5 is the only solution, since with x^2>0, clearly x>0 is not correct. For x^2>0, the domain of x is all real x. So, that leaves us with only one other bound: x>-15/2, which is -7 -1/2, so x=-3,5.

Perhaps I should have tested x=-3 before posting.

Can somebody explain to me how to solve, completely algabraically x^2>0? Normally if you raise each side to the power of 1/2, you get something like x=+/-a. But +0=0=-0...

 

[withoutaface]

yeah i know that but how the hell do u think of this stuff off the top of ur head?

I have absolutely no idea.
EDIT: I lied, yes I do, I've been doing way too much maths this weekend.

 

[Xayma]

And, taking it into consideration, I must revise my statement that x=5 is the only solution, since with x^2>0, clearly x>0 is not correct. For x^2>0, the domain of x is all real x. So, that leaves us with only one other bound: x>-15/2, which is -7 -1/2, so x=-3,5.

Perhaps I should have tested x=-3 before posting.

Can somebody explain to me how to solve, completely algabraically x^2>0? Normally if you raise each side to the power of 1/2, you get something like x=+/-a. But +0=0=-0...

No the domain is not all real x, it is all real except x=0.

 

[Iota]

'Tis true. *sigh*. Go away.

 

[kpq_sniper017]

that's the only expression on my face right now, and it just about sums up any words i can think of typing.