<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] $$(\log_{2}(x+1))^2=4\iff\log_{2}(x+1)=\pm2$$
[1]$$\log_{2}(x+1)=2,x+1=2^2,x_1=3$$
[2]$$\log_{2}(x+1)=-2,x+1=2^{-2},x_2=-3/4$$
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