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I'm having trouble finding the solution to this problem. I'm unsure how to go about solving the equation in regards to the exponent. Moving the 2 in front of the (ln x) would make the equation equate to 0=0 which is wrong. Any help would be much appreciated!

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If $f(x)=0$ then \begin{align} 0 &= (\ln x )^2 - 2 \ln x \\ \iff 0 &= u^2-2u\, \qquad (u=\ln x) \\ \iff 0 &= u(u-2) \\ \iff &u=0\, \text{or}\, u = 2 \end{align}

Can you take it from here?

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  • $\begingroup$ Yeah, that makes it much clearer thanks! $\endgroup$ – Matthew Bolster Aug 11 '17 at 9:45
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\begin{align*}f(x)=0&\iff\ln(x)^2=2\ln(x)\\&\iff\ln(x)=0\vee\ln(x)=2\\&\iff x=1\vee x=e^2.\end{align*}

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    $\begingroup$ @BjörnFriedrich Yes, I know that. And... ? $\endgroup$ – José Carlos Santos Aug 11 '17 at 9:11
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    $\begingroup$ @BjörnFriedrich Because $f(x)=\ln(x)^2-2\ln(x)$ and therefore $f(x)=0\iff\ln(x)^2=2\ln(x)$. What's the problem? $\endgroup$ – José Carlos Santos Aug 11 '17 at 9:13
  • $\begingroup$ @MatthewBolster See my solution below for a further explanation. $\endgroup$ – Kevin Aug 11 '17 at 9:44
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Hint:

$$(\ln x)^2 \ne \ln (x^2)$$

solve the equation $$ (\ln x)^2-2 \ln x=0 $$ with the substitution $\ln x=t$

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    $\begingroup$ @Kevin: Thank you. I edit the typo .. $\endgroup$ – Emilio Novati Aug 11 '17 at 8:54

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