Question

Show that:

−ln(x−x2−1−−−−−√)=ln(x+x2−1−−−−−√)

Answer 1

\[-\ln (x-\sqrt{x ^{2}-1})=\ln(x+\sqrt{x ^{2}-1})\]

Answer 2

I dont think this is true

Answer 3

how do i prove it?

Answer 4

plug in 2 on both sides

Answer 5

or 1 would be easier

Answer 6

err 1 is a solution

Answer 7

so plug in anything besides 1

Answer 8

ok

Answer 9

they cancel each other

Answer 10

\[-\ln (2-\sqrt{3})=\ln (2+\sqrt{3})\]

Answer 11

those are not the same thing

Answer 12

\(-\ln (2-\sqrt{3})=\ln (2+\sqrt{3})\\\implies \ln (\frac{1}{2-\sqrt{3}})=\ln (2+\sqrt{3})\\\implies \frac{1}{2-\sqrt{3}}=2+\sqrt{3}\)

Answer 13

hmmm http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItbG4oeC1zcXJ0KHheMi0xKSkiLCJjb2xvciI6IiMxRjJDQkEifSx7InR5cGUiOjAsImVxIjoibG4oeCtzcXJ0KHheMi0xKSkiLCJjb2xvciI6IiNGN0Y3MEEifSx7InR5cGUiOjEwMDAsIndpbmRvdyI6WyIwLjMzNDIzMzYwMDAwMDAwMDEzIiwiMy43NDIxMDU2MDAwMDAwMDA0IiwiLTAuMzkzMjE2IiwiMS43MDM5MzYwMDAwMDAwMDAzIl19XQ--


shows that they actually are

Answer 14

I don't believe it. But then again it is summer and I have been out of school for a month...

Answer 15

\(\large\color{black}{ \displaystyle -\ln(x-\sqrt{x^2-1})=\ln(x+\sqrt{x^2-1}) }\)


\(\large\color{black}{ \displaystyle \ln\left(\frac{1}{x-\sqrt{x^2-1}}\right)=\ln(x+\sqrt{x^2-1}) }\)


\(\large\color{black}{ \displaystyle \frac{1}{x-\sqrt{x^2-1}}=x+\sqrt{x^2-1} }\)


\(\large\color{black}{ \displaystyle1=\left(x+\sqrt{x^2-1} \right)\left(x-\sqrt{x^2-1}\right) }\)


\(\large\color{black}{ \displaystyle1=x^2-x\sqrt{x^2-1}+x\sqrt{x^2-1}-(x^2-1) }\)

Answer 16

and then simplify that...

Answer 17

Ok so why does it seem that math is broken? i.e. what i did with x=2?

Answer 18

wow I am out of it

Answer 19

I was just assuming 1/(2-sqrt{3}) < 1

Answer 20

that cant be a thing. you cant disprove it for x=2 and prove it for x= anything...

Answer 21

But it is true for x=2. I am just slow today...

Answer 22

\(\frac{1}{2-\sqrt{3}}=2+\sqrt{3}\) is true

Answer 23

@radicalgal the above process by @SolomonZelman is correct

Answer 24

and the graph, as show above as well, shows that both graph indeed are one and the same, thus both of them are equal indeed

Answer 25

but there are extraneous solutions somewhere when x<0.

Answer 26

besides these extraneous solutions, everything is correct I suppose...
Thanks jdoe0001

Answer 27

just to expand some ---> \(\bf 1=\left(x+\sqrt{x^2-1} \right)\left(x-\sqrt{x^2-1}\right) \implies 1=(x)^2-(\sqrt{x^2-1})^2
\\ \quad \\
1=x^2-(x^2-1)\implies 1=\cancel{x^2-x^2}+1\)

Answer 28

It is correct but not written correctly for a direct proof(the same way that it is not OK to manipulate both sides in a trig identity). But that depends on the level of proof expected....