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I want to understand how these equations have been solved:

Here's the first equation:

$$\log^2(x) + \frac{1}{2}\log(x^2)-2 = 0$$

$$= \log^2(x)+\log(x)-2=0 \text{ (What is this rule called here?)}$$ $$\Rightarrow u \equiv \log(x): u^2 + u-2 = 0 \text{ (Substitution)}$$ $$\Rightarrow u^2 + u - 2 = (u+2)(u-1) \Rightarrow u_1 = -2, u_2 = 1 $$ $$x_1 =e^{-2}, x_2=e^1 \text{ (How do we know that we have to use e here?)}$$

Here's the second equation:

$$\log_2(x)+\log_2(x+2)-2 = 0$$ $$=\log_2(x^2+2x)-2= 0 \text{ (Can we just multiply it even though it is a sum? } x \cdot (x+2))$$ $$\Leftrightarrow \log_2(x^2+2x) = 2$$ $$\Leftrightarrow x^2 +2x = 2^2 = 4 \text{ }$$ $$\Rightarrow x_{1,2} = \frac{1}{2} \big(-2 \pm \sqrt{4-4\cdot(-8)}\big) = -1 \pm \sqrt{1+8}$$ $$\Rightarrow x_1=2, x_2 = -4$$

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    $\begingroup$ $(1)\quad\log_a b ^x \implies \color{red}{x}\log_a b \quad (2.)$ Since no base is mentioned , we generally consider the natural logarithm $\quad (3.)\quad \log_a{xy} \implies \log_a\color{red}{x} + \log_a\color{blue}{y}$ $\endgroup$ – The Demonix _ Hermit Dec 6 '19 at 10:44
  • $\begingroup$ @TheDemonix_Hermit Thanks a lot. Can you also explain how we get to $x^2+2x = 2^3$? $\endgroup$ – Retarded Programmer Dec 6 '19 at 10:50
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    $\begingroup$ There's an error: it should be $2^2$ on the r.h.s. $\endgroup$ – Bernard Dec 6 '19 at 10:59
  • $\begingroup$ @Bernard Thanks! $\endgroup$ – Retarded Programmer Dec 6 '19 at 11:01
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The questions's been answered in the comments

$$\log_ab^x \Rightarrow x \log b$$ $$\log_axy \Rightarrow \log_a x + \log_a y$$

and an error on the r.h.s

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