anonymous
  • anonymous
Show that: −ln(x−x2−1−−−−−√)=ln(x+x2−1−−−−−√)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[-\ln (x-\sqrt{x ^{2}-1})=\ln(x+\sqrt{x ^{2}-1})\]
zzr0ck3r
  • zzr0ck3r
I dont think this is true
anonymous
  • anonymous
how do i prove it?

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zzr0ck3r
  • zzr0ck3r
plug in 2 on both sides
zzr0ck3r
  • zzr0ck3r
or 1 would be easier
zzr0ck3r
  • zzr0ck3r
err 1 is a solution
zzr0ck3r
  • zzr0ck3r
so plug in anything besides 1
anonymous
  • anonymous
ok
anonymous
  • anonymous
they cancel each other
anonymous
  • anonymous
\[-\ln (2-\sqrt{3})=\ln (2+\sqrt{3})\]
zzr0ck3r
  • zzr0ck3r
those are not the same thing
zzr0ck3r
  • zzr0ck3r
\(-\ln (2-\sqrt{3})=\ln (2+\sqrt{3})\\\implies \ln (\frac{1}{2-\sqrt{3}})=\ln (2+\sqrt{3})\\\implies \frac{1}{2-\sqrt{3}}=2+\sqrt{3}\)
jdoe0001
  • jdoe0001
hmmm http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItbG4oeC1zcXJ0KHheMi0xKSkiLCJjb2xvciI6IiMxRjJDQkEifSx7InR5cGUiOjAsImVxIjoibG4oeCtzcXJ0KHheMi0xKSkiLCJjb2xvciI6IiNGN0Y3MEEifSx7InR5cGUiOjEwMDAsIndpbmRvdyI6WyIwLjMzNDIzMzYwMDAwMDAwMDEzIiwiMy43NDIxMDU2MDAwMDAwMDA0IiwiLTAuMzkzMjE2IiwiMS43MDM5MzYwMDAwMDAwMDAzIl19XQ-- shows that they actually are
zzr0ck3r
  • zzr0ck3r
I don't believe it. But then again it is summer and I have been out of school for a month...
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle -\ln(x-\sqrt{x^2-1})=\ln(x+\sqrt{x^2-1}) }\) \(\large\color{black}{ \displaystyle \ln\left(\frac{1}{x-\sqrt{x^2-1}}\right)=\ln(x+\sqrt{x^2-1}) }\) \(\large\color{black}{ \displaystyle \frac{1}{x-\sqrt{x^2-1}}=x+\sqrt{x^2-1} }\) \(\large\color{black}{ \displaystyle1=\left(x+\sqrt{x^2-1} \right)\left(x-\sqrt{x^2-1}\right) }\) \(\large\color{black}{ \displaystyle1=x^2-x\sqrt{x^2-1}+x\sqrt{x^2-1}-(x^2-1) }\)
SolomonZelman
  • SolomonZelman
and then simplify that...
zzr0ck3r
  • zzr0ck3r
Ok so why does it seem that math is broken? i.e. what i did with x=2?
zzr0ck3r
  • zzr0ck3r
wow I am out of it
zzr0ck3r
  • zzr0ck3r
I was just assuming 1/(2-sqrt{3}) < 1
zzr0ck3r
  • zzr0ck3r
that cant be a thing. you cant disprove it for x=2 and prove it for x= anything...
zzr0ck3r
  • zzr0ck3r
But it is true for x=2. I am just slow today...
zzr0ck3r
  • zzr0ck3r
\(\frac{1}{2-\sqrt{3}}=2+\sqrt{3}\) is true
jdoe0001
  • jdoe0001
@radicalgal the above process by @SolomonZelman is correct
jdoe0001
  • 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
SolomonZelman
  • SolomonZelman
but there are extraneous solutions somewhere when x<0.
SolomonZelman
  • SolomonZelman
besides these extraneous solutions, everything is correct I suppose... Thanks jdoe0001
jdoe0001
  • jdoe0001
just to expand some ---> \(\bf 1=\left(x+\sqrt{x^2-1} \right)\left(x-\sqrt{x^2-1}\right) \implies 1=(x)^2-(\sqrt{x^2-1})^2 \\ \quad \\ 1=x^2-(x^2-1)\implies 1=\cancel{x^2-x^2}+1\)
zzr0ck3r
  • 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....

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