anonymous 5 years ago What is the solution to the equation (lny)^2 - 2ln(y^5) = 11, and how do you get the derivative of the equation with respect to y?

1. anonymous

wait, why are you trying to do a derivative of the equation? this is a simple quadratic equation.

2. anonymous

Like as in -b(+-) sqrt b^2-4ac over 2a?

3. anonymous

what is $\ln a ^{b}$ ?

4. anonymous

Oh, so I can just plug all of that into the quadratic equation to get my solution and then take the derivative?

5. anonymous

No, you dont need to take any derivatives at all.

6. anonymous

I'm so confused.

7. anonymous

just answer my question. what is $\ln a ^{b}$

8. anonymous

a^b?

9. anonymous

okay do you know about logarithms?

10. anonymous

A little. I had calculus last semester...and I'm having to refresh everything...lol. If I took the derivative of the equation you just gave me, then I would get a^b right?

11. anonymous

I didn't ask for the derivative of log a^b. I just asked what log a^b was.

12. anonymous

okay log a^b = b log a

13. anonymous

Oh! Yes.

14. anonymous
15. anonymous

do you know what logarithm means though?

16. anonymous

don't worry, thats a side question.

17. anonymous

Holy crap! So, really the problem should look like...$((lny)^2)/(2lny^5)$

18. anonymous

NO! how did you get THAT?

19. anonymous

(lny)^2 - 2lny^5 = 11 or (lny)^2 - 10 lny -11 = 0 say ln y = x x^2-10x-11 =0 solve for x, substitute in ln y

20. anonymous

got it?

21. anonymous

need feedback to know you got it.

22. anonymous

Ohhh OKay I got it, I'm sorry for the delay. I was making dinner. That makes sense. See, I started dividing because I figured if you have ln (2-3) then it's ln 2/ln 3. Idk. I don't know why I thought that.