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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln (x\sqrt{x ^{2}1})=\ln(x+\sqrt{x ^{2}1})\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I dont think this is true

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1plug in 2 on both sides

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1so plug in anything besides 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they cancel each other

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln (2\sqrt{3})=\ln (2+\sqrt{3})\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1those are not the same thing

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1\(\ln (2\sqrt{3})=\ln (2+\sqrt{3})\\\implies \ln (\frac{1}{2\sqrt{3}})=\ln (2+\sqrt{3})\\\implies \frac{1}{2\sqrt{3}}=2+\sqrt{3}\)

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0hmmm http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiItbG4oeC1zcXJ0KHheMi0xKSkiLCJjb2xvciI6IiMxRjJDQkEifSx7InR5cGUiOjAsImVxIjoibG4oeCtzcXJ0KHheMi0xKSkiLCJjb2xvciI6IiNGN0Y3MEEifSx7InR5cGUiOjEwMDAsIndpbmRvdyI6WyIwLjMzNDIzMzYwMDAwMDAwMDEzIiwiMy43NDIxMDU2MDAwMDAwMDA0IiwiLTAuMzkzMjE2IiwiMS43MDM5MzYwMDAwMDAwMDAzIl19XQ shows that they actually are

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I don't believe it. But then again it is summer and I have been out of school for a month...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \ln(x\sqrt{x^21})=\ln(x+\sqrt{x^21}) }\) \(\large\color{black}{ \displaystyle \ln\left(\frac{1}{x\sqrt{x^21}}\right)=\ln(x+\sqrt{x^21}) }\) \(\large\color{black}{ \displaystyle \frac{1}{x\sqrt{x^21}}=x+\sqrt{x^21} }\) \(\large\color{black}{ \displaystyle1=\left(x+\sqrt{x^21} \right)\left(x\sqrt{x^21}\right) }\) \(\large\color{black}{ \displaystyle1=x^2x\sqrt{x^21}+x\sqrt{x^21}(x^21) }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1and then simplify that...

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Ok so why does it seem that math is broken? i.e. what i did with x=2?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1I was just assuming 1/(2sqrt{3}) < 1

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1that cant be a thing. you cant disprove it for x=2 and prove it for x= anything...

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1But it is true for x=2. I am just slow today...

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1\(\frac{1}{2\sqrt{3}}=2+\sqrt{3}\) is true

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0@radicalgal the above process by @SolomonZelman is correct

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0and 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
 one year ago
Best ResponseYou've already chosen the best response.1but there are extraneous solutions somewhere when x<0.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1besides these extraneous solutions, everything is correct I suppose... Thanks jdoe0001

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0just to expand some > \(\bf 1=\left(x+\sqrt{x^21} \right)\left(x\sqrt{x^21}\right) \implies 1=(x)^2(\sqrt{x^21})^2 \\ \quad \\ 1=x^2(x^21)\implies 1=\cancel{x^2x^2}+1\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1It 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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