Question

lnx+ln(x−2)=4 for x

Answer 1

start with
\[\ln(x^2-2x)=4\] then rewrite in exponential form

Answer 2

how do you do that?

Answer 3

\[\ln(\spadesuit)=\heartsuit\iff e^{\heartsuit}=\spadesuit\]

Answer 4

so e^4 = (x^2 - 2x)

Answer 5

yes

Answer 6

then solve for \(x\)

Answer 7

probably easiest to complete the square for this one, since 2 is even
you could use the quadratic formula but it would be more work

Answer 8

does the equation first become 4 = x^2 -2x ?

Answer 9

or did i do this completely wrong?

Answer 10

@satellite73

Answer 11

x^2-2x=e^4

Answer 12

i know but what would you do after with the e^4 ? @freckles

Answer 13

I would solve the quadratic equation by using the quadratic formula.

Answer 14

You have ax^2+bx+c=0

Answer 15

\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 16

c would be -4?

Answer 17

c would be -e^4
you have x^2-2x=e^4
not x^2-2x=4

Answer 18

oh okay that's where i went wrong, i thought you had to change the e to something

Answer 19

so i got -2 +/- (-2+ e^4 )^.5

Answer 20

b^2=4 not -2

Answer 21

and i assume you are just writing the top

Answer 22

\[x^2-2x=e^4 \\ x^2-2x-e^4=0 \\ a=1 \\ b=-2 \\ c=-e^4 \\ x=\frac{-(-2) \pm \sqrt{ (-2)^2-4(1)(-e^4)}}{2(1)}\]

Answer 23

I just plugged into the formula

Answer 24

oh i simplified 4 +/- (-2 + e^4)/ 2

Answer 25

wait that's not right

Answer 26

you put b^2 i the wrong place

Answer 27

is it 2 +/- (4 + 4e^4)^.5 / 2

Answer 28

looks looks a lot better

Answer 29

this can be simplified though

Answer 30

\[x^2-2x=e^4 \\ x^2-2x-e^4=0 \\ a=1 \\ b=-2 \\ c=-e^4 \\ x=\frac{-(-2) \pm \sqrt{ (-2)^2-4(1)(-e^4)}}{2(1)} \\ x=\frac{2 \pm \sqrt{4+4e^4}}{2} \\ x=\frac{2 \pm \sqrt{4} \sqrt{1+e^4}}{2}\]

Answer 31

so it' (1+e^4)^1/2 + 1

Answer 32

\[x^2-2x=e^4 \\ x^2-2x-e^4=0 \\ a=1 \\ b=-2 \\ c=-e^4 \\ x=\frac{-(-2) \pm \sqrt{ (-2)^2-4(1)(-e^4)}}{2(1)} \\ x=\frac{2 \pm \sqrt{4+4e^4}}{2} \\ x=\frac{2 \pm \sqrt{4} \sqrt{1+e^4}}{2} \\ x=\frac{2 \pm 2 \sqrt{1+e^4}}{2}=\frac{2(1 \pm \sqrt{1+e^4})}{2}=1\pm \sqrt{1+e^4}\]

Answer 33

yeah but i think you can't take the negative of a log so it's just the positive one

Answer 34

or something like that

Answer 35

yes you should check both solutions

Answer 36

thought 1+sqrt(1+e^4) is obviously positive

Answer 37

sqrt(1+e^4) is more than 1 i think
1-u is negative when u>1

Answer 38

and yeah sqrt(1+e^4) is 7.something
1-7.something is definitely negative

Answer 39

ok thank you so much!!