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$(\ln(x) - 1)^2 = 4$

My approach was to root both sides:

$\ln(x) - 1 = 2 $

$\therefore e^2 = x - 1$

$\therefore x = e^2 + 1$

The answers are $e^3$ and $e^{-1}$ that are found from expanding the parens. I was wondering if someone could please help explain to me why my answer is invalid.

Thanks

Robert Z
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Henry Cooper
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3 Answers3

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Hints:

  • The number $2$ is not the only one whose square is $4$.

$a^2 = 4 \iff a = 2 \vee a = \ldots$

  • $\ln x -1 \ne \ln\left(x-1\right)$, so be careful when solving for $x$.

$\ln x - 1 = y \iff \ln x = y +1 \iff x = \ldots$

Try again? Ask for help if you don't get to the given answers.

StackTD
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Assuming you mean $$\ln(x-1)^2 = 4$$

Then \begin{align} \ln (x-1) &= \pm 2 \\ \implies x-1 = e^{\pm 2}\\ \implies x &= 1+e^{\pm 2} \end{align}

However, if you mean $$(\ln(x)-1)^2 = 4$$ Then \begin{align} \ln (x)-1 &= \pm 2 \\ \implies \ln x &= 1\pm 2 \\ \implies x &= e^{1\pm 2} \end{align}

Obviously I leave it to you to determine if there is only one solution.

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$\ln x - 1 = 2 \implies e^2=x-1$ is false. At this stage instead use $\ln x - 1 = 2 \implies \ln x = 3$ and go from there to get the answer you're expecting.

CyclotomicField
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    What a fool I was! This was the answer that set me straight. I can’t vote up but have accepted. Thank you kindly – Henry Cooper Oct 16 '18 at 14:48
  • @HenryCooper it's an easy mistake to make because it's notational not mathematical. It doesn't indicate any serious misunderstanding of the material so I think you're doing fine. Everyone makes simple errors like this now and then, especially if there is time pressure. – CyclotomicField Oct 16 '18 at 16:28