How do I simplify ln(1-y^2)=x^2+C.
I can solve for see, but I'm trying to figure out how to remove the ln

Question

How do I simplify ln(1-y^2)=x^2+C.
I can solve for see, but I'm trying to figure out how to remove the ln

owlcoffee · Answer

I'll give you a equality:
$$e ^{Lnx}=x$$
or, you can apply the definition of logarithm:
$$\log_{a} B=C <=> a^C=B$$
So applying it:
$$\ln(1-y^2)=x^2+c$$
$$e ^{x^2+c}=1-y^2$$

anonymous · Answer

I forgot, the 1-y^2 is under a square root so to remove that would I get 1-y^2=(e^c *e^x^2)^2

owlcoffee · Answer

You are correct.

anonymous · Answer

Okay, thank you.

javalos01 · Answer

r u questioning us or is it ur question...just asking

parthkohli · Answer

$$\ln (x^m) = m \ln(x)$$

javalos01 · Answer

true

parthkohli · Answer

$$\ln(1 - y^2) = 2x^2 + 2C$$Just a question: does C denote a constant here?

javalos01 · Answer

good job @debins33  n' @ParthKohli

anonymous · Answer

Yes, c is a constant

parthkohli · Answer

Then$$\ln(1 - y^2) = 2x^2 + C$$

parthkohli · Answer

$$1 - y^2 = e^{2x^2 + C}  = Ce^{2x^2}$$

javalos01 · Answer

yea its that one

anonymous · Answer

So then y=sqrt(-Ce^2x^2)+1?

parthkohli · Answer

Yeah, I guess so, but the + 1 is inside the sqrt.

anonymous · Answer

Okay, that's right. Thank you

parthkohli · Answer

Just for the record, I wrote $e^{2x^2 + C}$ as $Ce^{2x^2}$ since $e^C$ is also a constnat.

parthkohli · Answer

constant*

This lag is annoying.

anonymous · Answer

OKay