Question

Please help! Solve for x.
lnx(x^2-3x-4)+lnx(x^2-9x+20)=0

Answer 1

divide both sides by lnx?

Answer 2

@Mertsj you up to helping me out? the other guy from earlier said he didnt know how to do it without a calculator

Answer 3

why don't you divide both sides by lnx? unless you meant to write something different

Answer 4

im not sure you can do that. (oh i wasnt talking about you btw)

Answer 5

if what you meant is lnx times (x^2-3x-4) then you can factor it out

Answer 6

and then?

Answer 7

there are two ways you can approach this
lnx(x^2-3x-4)+lnx(x^2-9x+20)=0
either divide everything by lnx

(x^2-3x-4)+(x^2-9x+20)=0

or lnx(x^2-3x-4)=-lnx(x^2-9x+20)
and raise to the e to get
x(x^2-3x-4)=-x(x^2-9x+20)

Answer 8

ok one second let me try the first way

Answer 9

um nevermind idk how to do it -_-

Answer 10

(x^2-3x-4)+(x^2-9x+20)=0
2x^2-12x+16=0
divide by 2
x^2-6x+8=0

factorise to get (x-4)(x-2)=0
x=2 or x=4

Answer 11

ok that makes sense but the answer key has x=1 as an answer as well.

Answer 12

in that case i think i'm misunderstanding the question. could you type it using the equation signs?

Answer 13

\[\ln x(x^2-3x-4)+lnx(x^2-9x+20)=0\]

Answer 14

2 and 4 are answers as well, but one is also

Answer 15

ah of course it is! we have factored out lnx which is also a solution. silly me!
lnx = 0
e ^ lnx = e ^ 0
x = 1

Answer 16

ahh ok! thanks for your time!

Answer 17

no problem ;)

Answer 18

\[xy _{1}+xy _{1}y^2-3=5y _{1}+xy \] solve for \[y _{1}\]