The Student Room Group

Logs

Solve lnx=6/(5-lnx)

I know the answer is e^2 or e^3 but I don't know how to get there. Any hints?
Reply 1
Multiply the equation by 5-lnx and you will have a quadratic in lnx.
Reply 2
5ln^2x-lnx=6 ?
I tried solving that with no luck, I'll go over my working again though
Reply 3
Oh, basic factorising mistake, thanks
(5lnx)(lnx)=6(5-\ln x)(\ln x) = 6
5lnx(lnx)2=65\ln x - (\ln x)^2 = 6

Let y = ln x

y25y+6=0y^2 - 5y +6 = 0

y=3,y=2y=3, y=2

lnx=3,x=e3\ln x = 3 , x = e^3
lnx=2,x=e2\ln x = 2 , x = e^2


Now I get neg rep for helping out ?
(edited 13 years ago)
Original post by sdgvrgverg
5ln^2x-lnx=6 ?
I tried solving that with no luck, I'll go over my working again though


Nope, it should be ln^2x - 5lnx + 6 = 0, then you could take lnx = y and you'll have y^2 - 5y + 6 = 0 which you could then factorise to give (y-2)(y-3) therefore the solutions are y = 3 and y = 2 which means lnx = 2 and lnx = 3 as y = lnx, then you do:

lnx = 2
x = e^2

lnx = 3
x = e^3

=D
Reply 6
Original post by CookieGhoul
(5lnx)(lnx)=6(5-\ln x)(\ln x) = 6
5lnx(lnx)2=65\ln x - (\ln x)^2 = 6

Let y = ln x

y25y+6=0y^2 - 5y +6 = 0

y=3,y=2y=3, y=2

lnx=3,x=e3\ln x = 3 , x = e^3
lnx=2,x=e2\ln x = 2 , x = e^2


Now I get neg rep for helping out ?


Because you're not supposed to give him the full solution.

Quick Reply

Latest