## anonymous one year ago Show that: −ln(x−x2−1−−−−−√)=ln(x+x2−1−−−−−√)

1. anonymous

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

2. zzr0ck3r

I dont think this is true

3. anonymous

how do i prove it?

4. zzr0ck3r

plug in 2 on both sides

5. zzr0ck3r

or 1 would be easier

6. zzr0ck3r

err 1 is a solution

7. zzr0ck3r

so plug in anything besides 1

8. anonymous

ok

9. anonymous

they cancel each other

10. anonymous

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

11. zzr0ck3r

those are not the same thing

12. zzr0ck3r

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

13. jdoe0001
14. zzr0ck3r

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

15. SolomonZelman

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

16. SolomonZelman

and then simplify that...

17. zzr0ck3r

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

18. zzr0ck3r

wow I am out of it

19. zzr0ck3r

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

20. zzr0ck3r

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

21. zzr0ck3r

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

22. zzr0ck3r

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

23. jdoe0001

@radicalgal the above process by @SolomonZelman is correct

24. jdoe0001

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

25. SolomonZelman

but there are extraneous solutions somewhere when x<0.

26. SolomonZelman

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

27. jdoe0001

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

28. zzr0ck3r

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....