Question

log x+log(x-21)=2 me ajudar aí pessoal

Answer 1

Lembrete:

$\cdot \log _b\left(a\right)\:\:\rightarrow \:\:a\ \textgreater \ 0$
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$\log \left(x\right)+\log \left(x-21\right)=2\\\log \left(x\left(x-21\right)\right)=2\\\log \left(x^2-21x\right)=2\\\log \left(x^2-21x\right)=\log \left(100\right)\\x^2-21x=100\\x^2-21x-100=0\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\x=\frac{21\pm \sqrt{21^2-4\cdot 1\cdot -\left(100\right)}}{2\cdot 1}\\x=\frac{21\pm \sqrt{441+400}}{2}\\x=\frac{21\pm \sqrt{841}}{2}\\x=\frac{21\pm 29}{2}\\\\\boxed{\bold{x_1=\frac{21+29}{2}=25}}\\\\\boxed{\bold{x_2=\frac{21-29}{2}=-4\:\left(FALSO\right)}}$

Respostas: S={25}