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I would like to solve the equation: $2\ln(y) = x$

First way: $$2\ln y = x \Leftrightarrow e^{2\ln y}=e^x \Leftrightarrow y^2=e^x\Rightarrow y= \pm \sqrt{e^x}$$ Second way: $$2\ln y = x \Leftrightarrow \ln y= \frac x2 \Leftrightarrow e^{\ln y}=e^{ \frac x2} \Leftrightarrow y=e^{ \frac x2} \Leftrightarrow y= \sqrt{e^x}$$

Why do we get an extra root in the first solution? Which solution is the correct one?

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$\ln y$ is defined only for $y>0$ so you have to choose the positive square root in the first method.

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  • $\begingroup$ But why do I get an extraneous root in the first case? I apply the function $e^x$ on both sides which is 1-1, hence the double arrow that I used. $\endgroup$ – Nick Sep 19 '19 at 12:14
  • $\begingroup$ @Nick Start with $y=-\sqrt {e^{x}}$ and work backwards.See where you get into trouble. $\endgroup$ – Kavi Rama Murthy Sep 19 '19 at 12:17
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Have you already checked your result with a numeric example?

Let's say $x = 2$ and thus your solution for y = -2.717882 and + 2.717882.

but the base of the natural logarithm should always be positive. For a good explanation why, see this post:

Why must the base of a logarithm be a positive real number not equal to 1?

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