Question

ln(x)+ln(x-1)=2
solve for x

Answer 1

use product rule for log on left hand side

Answer 2

then write an equivalent exponential form

Answer 3

\[\ln(a)=\log_e(a) \\ \text{ so if you have } \log_e(a)=y \text{ then } e^{y}=a\]

Answer 4

still need some help with it

Answer 5

have you used the product rule yet?

Answer 6

can you show me what you have after doing that please

Answer 7

lnx(x-1)=2

Answer 8

\[\ln(x(x-1))=2 \\ \ln(x^2-x)=2 \\ \text{ now can you write this \in equivalent exponential form } \\ \text{ using what I mentioned }\]

Answer 9

so would you end up with e^2=x^2-x?

Answer 10

that is right

Answer 11

you have a quadratic equation

Answer 12

\[e^2=x^2-x \\ 0=x^2-x-e^2 \\ x^2-x-e^2=0 \]

Answer 13

\[\text{ recall if } ax^2+bx+c=0 \text{ then } x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 14

make sure your final answers fit the domain of the original equation

Answer 15

it won't except any way I put it in

Answer 16

how are you putting the answer in ?

Answer 17

exact form ?
approximation?
and can i also know what you have put in

Answer 18

and did you make sure you answers fit the domain of the original equation ?

Answer 19

I put 1+sqrt(1-4e^2)/2

Answer 20

I see one problem

Answer 21

why do you have a - between 1 and 4e^2 ?

Answer 22

because it's b^2-4ac underneath the squareroot

Answer 23

\[x^2-x-e^2=0 \\ x=\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-e^2)}}{2(1)} \\ x=\frac{-1 \pm \sqrt{1+4 e^2}}{2} \\ \text{ the solution we have is } x=\frac{1+\sqrt{1+4e^2}}{2}\]

Answer 24

yes but a negative times a negative isn't negative

Answer 25

it is positive

Answer 26

you have -4 *-e^2 this is not -4e^2
it is 4e^2

Answer 27

opps, I missed that thanks

Answer 28

also I made a type-o above but it was corrected in next line I had above

Answer 29

\[x^2-x-e^2=0 \\ x=\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-e^2)}}{2(1)} \\ x=\frac{1 \pm \sqrt{1+4 e^2}}{2} \\ \text{ the solution we have is } x=\frac{1+\sqrt{1+4e^2}}{2}\]*