determine o conjunto soluçao de log12(x^2-x)=1
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Usando a definição de logaritmo
![a^x=b \\ \\ log_{a} b =x a^x=b \\ \\ log_{a} b =x](https://tex.z-dn.net/?f=a%5Ex%3Db+%5C%5C++%5C%5C++log_%7Ba%7D+b+%3Dx)
![12^1=x^2-x \\ \\ x^2-x-12=0 12^1=x^2-x \\ \\ x^2-x-12=0](https://tex.z-dn.net/?f=12%5E1%3Dx%5E2-x+%5C%5C++%5C%5C+x%5E2-x-12%3D0)
Equação de 2º grau resolvemos pela fórmula de Bhaskara, anexa
a= 1 b=-1 c=-12
![delta = (-1)^2-4.1.(-12) = 1+48 = 49 \\ \\ \sqrt{delta} = \sqrt{49} =7 \\ \\ x_{1} = \frac{-(-1)+7}{2.1} = \frac{8}{2} =4 \\ \\ x_{2} = \frac{-(-1)-7}{2.1} = \frac{-6}{2} =-3 delta = (-1)^2-4.1.(-12) = 1+48 = 49 \\ \\ \sqrt{delta} = \sqrt{49} =7 \\ \\ x_{1} = \frac{-(-1)+7}{2.1} = \frac{8}{2} =4 \\ \\ x_{2} = \frac{-(-1)-7}{2.1} = \frac{-6}{2} =-3](https://tex.z-dn.net/?f=delta+%3D+%28-1%29%5E2-4.1.%28-12%29+%3D+1%2B48+%3D+49+%5C%5C++%5C%5C++%5Csqrt%7Bdelta%7D+%3D+%5Csqrt%7B49%7D+%3D7+%5C%5C++%5C%5C++x_%7B1%7D+%3D+%5Cfrac%7B-%28-1%29%2B7%7D%7B2.1%7D+%3D+%5Cfrac%7B8%7D%7B2%7D+%3D4+%5C%5C++%5C%5C++x_%7B2%7D+%3D+%5Cfrac%7B-%28-1%29-7%7D%7B2.1%7D+%3D+%5Cfrac%7B-6%7D%7B2%7D+%3D-3)
S={-3,4}
Equação de 2º grau resolvemos pela fórmula de Bhaskara, anexa
a= 1 b=-1 c=-12
S={-3,4}
Anexos:
![](https://pt-static.z-dn.net/files/d4c/443422d55c9bcc8264ff2628920f8682.jpg)
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