<s>[INST] Solve$(log_{2}(x+1))^2=4$ $$(log_{2}(x+1))^2=4$$
$$log_{2}(x+1)*log_{2}(x+1)=log_{2}16$$
$$x^{2}+2x-15=0$$
$$(x+1)*(x+1)=16$$
$$x^{2}+2x+1=16$$
$$x^{2}+2x-15=0$$
$$(x+5)(x-3)=0$$
$$x_1=-5; x_2=3$$
The solution is only $x_1=3$. Is this correct?
 [/INST] Your solution is incorrect. First note that
$$\log_2\left(a^2 \right) \neq \left(\log_2(a) \right)^2$$
In your case, we have
$$\left(\log_2(x+1) \right)^2 = 4 \implies \log_2(x+1) = \pm2 \implies x+1 = 2^{\pm2} \implies x+1 = 4 \text{ or }x+1= \dfrac14$$
Hence,
$$x=3 \text{ or }x=-\dfrac34$$
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