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 Post subject: ln square root
PostPosted: Thu, 1 Mar 2012 22:32:42 UTC 
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Joined: Thu, 1 Mar 2012 22:28:45 UTC
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I need to solve for x in this problem:

ln(3sqrtx)=sqrt(ln(x))


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 Post subject: Re: ln square root
PostPosted: Thu, 1 Mar 2012 23:31:29 UTC 
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Joined: Mon, 19 May 2003 19:55:19 UTC
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Hello, orange23!

Is there a typo? . . . I'm getting a complex answer.


Quote:
\ln(3\sqrt{x}) \:=\:\sqrt{\ln x}

We have: .\ln 3 + \ln \sqrt{x} \:=\:\sqrt{\ln x}

. . . . . . . \ln 3 + \frac{1}{2}\ln x \:=\:\sqrt{\ln x} \quad\Rightarrow\quad 2\ln 3 + \ln x \:=\:2\sqrt{\ln x}


Square both sides: .4(\ln 3)^2 + 4\ln 3\ln x + (\ln x)^2 \:=\:4\ln x

n . . . . . . . . (\ln x)^2 + 4(\ln 3 - 1)\ln x + 4(\ln 3)^2 \:=\:0 . . . a quadratic!


\text{Quadratic Formula: }\:\ln x \;=\;\dfrac{-4(\ln3-1) \pm \sqrt{16(\ln3 -1)^2 - 16(\ln 3)^2}}{2}

. . . . . . . . . . . . . . . . \ln x \;=\;-2(\ln 3 - 1) \pm 2\sqrt{1-2\ln 3}


But (1 - 2\ln 3) is negative . . . There are no real roots.



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 Post subject: Re: ln square root
PostPosted: Fri, 2 Mar 2012 00:14:52 UTC 
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there is no typo... this sounds good to me, thanks for all your help, I couldn't have gotten anywhere close to this answer!


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