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anonymous

  • one year ago

Show that: −ln(x−x2−1−−−−−√)=ln(x+x2−1−−−−−√)

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  1. anonymous
    • one year ago
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    \[-\ln (x-\sqrt{x ^{2}-1})=\ln(x+\sqrt{x ^{2}-1})\]

  2. zzr0ck3r
    • one year ago
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    I dont think this is true

  3. anonymous
    • one year ago
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    how do i prove it?

  4. zzr0ck3r
    • one year ago
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    plug in 2 on both sides

  5. zzr0ck3r
    • one year ago
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    or 1 would be easier

  6. zzr0ck3r
    • one year ago
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    err 1 is a solution

  7. zzr0ck3r
    • one year ago
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    so plug in anything besides 1

  8. anonymous
    • one year ago
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    ok

  9. anonymous
    • one year ago
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    they cancel each other

  10. anonymous
    • one year ago
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    \[-\ln (2-\sqrt{3})=\ln (2+\sqrt{3})\]

  11. zzr0ck3r
    • one year ago
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    those are not the same thing

  12. zzr0ck3r
    • one year ago
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    \(-\ln (2-\sqrt{3})=\ln (2+\sqrt{3})\\\implies \ln (\frac{1}{2-\sqrt{3}})=\ln (2+\sqrt{3})\\\implies \frac{1}{2-\sqrt{3}}=2+\sqrt{3}\)

  13. zzr0ck3r
    • one year ago
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    I don't believe it. But then again it is summer and I have been out of school for a month...

  14. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle -\ln(x-\sqrt{x^2-1})=\ln(x+\sqrt{x^2-1}) }\) \(\large\color{black}{ \displaystyle \ln\left(\frac{1}{x-\sqrt{x^2-1}}\right)=\ln(x+\sqrt{x^2-1}) }\) \(\large\color{black}{ \displaystyle \frac{1}{x-\sqrt{x^2-1}}=x+\sqrt{x^2-1} }\) \(\large\color{black}{ \displaystyle1=\left(x+\sqrt{x^2-1} \right)\left(x-\sqrt{x^2-1}\right) }\) \(\large\color{black}{ \displaystyle1=x^2-x\sqrt{x^2-1}+x\sqrt{x^2-1}-(x^2-1) }\)

  15. SolomonZelman
    • one year ago
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    and then simplify that...

  16. zzr0ck3r
    • one year ago
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    Ok so why does it seem that math is broken? i.e. what i did with x=2?

  17. zzr0ck3r
    • one year ago
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    wow I am out of it

  18. zzr0ck3r
    • one year ago
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    I was just assuming 1/(2-sqrt{3}) < 1

  19. zzr0ck3r
    • one year ago
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    that cant be a thing. you cant disprove it for x=2 and prove it for x= anything...

  20. zzr0ck3r
    • one year ago
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    But it is true for x=2. I am just slow today...

  21. zzr0ck3r
    • one year ago
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    \(\frac{1}{2-\sqrt{3}}=2+\sqrt{3}\) is true

  22. jdoe0001
    • one year ago
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    @radicalgal the above process by @SolomonZelman is correct

  23. jdoe0001
    • one year ago
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    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

  24. SolomonZelman
    • one year ago
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    but there are extraneous solutions somewhere when x<0.

  25. SolomonZelman
    • one year ago
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    besides these extraneous solutions, everything is correct I suppose... Thanks jdoe0001

  26. jdoe0001
    • one year ago
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    just to expand some ---> \(\bf 1=\left(x+\sqrt{x^2-1} \right)\left(x-\sqrt{x^2-1}\right) \implies 1=(x)^2-(\sqrt{x^2-1})^2 \\ \quad \\ 1=x^2-(x^2-1)\implies 1=\cancel{x^2-x^2}+1\)

  27. zzr0ck3r
    • one year ago
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    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....

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