Respuestas
Respuesta dada por:
2
Expandimos el logaritmo
![\log _{10}\left(3\right)x^2+\log _{10}\left(3\right)\left(x-6\right) \log _{10}\left(3\right)x^2+\log _{10}\left(3\right)\left(x-6\right)](https://tex.z-dn.net/?f=%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x%5E2%2B%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%5Cleft%28x-6%5Cright%29)
![=x\log _{10}\left(3\right)-6\log _{10}\left(3\right) =x\log _{10}\left(3\right)-6\log _{10}\left(3\right)](https://tex.z-dn.net/?f=%3Dx%5Clog+_%7B10%7D%5Cleft%283%5Cright%29-6%5Clog+_%7B10%7D%5Cleft%283%5Cright%29)
![=\log _{10}\left(3\right)x^2+x\log _{10}\left(3\right)-6\log _{10}\left(3\right) =\log _{10}\left(3\right)x^2+x\log _{10}\left(3\right)-6\log _{10}\left(3\right)](https://tex.z-dn.net/?f=%3D%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x%5E2%2Bx%5Clog+_%7B10%7D%5Cleft%283%5Cright%29-6%5Clog+_%7B10%7D%5Cleft%283%5Cright%29)
Nos da una ecuación cuadrática
![x^2\log _{10}\left(3\right)+x\log _{10}\left(3\right)-6\log _{10}\left(3\right)=0 x^2\log _{10}\left(3\right)+x\log _{10}\left(3\right)-6\log _{10}\left(3\right)=0](https://tex.z-dn.net/?f=x%5E2%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%2Bx%5Clog+_%7B10%7D%5Cleft%283%5Cright%29-6%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%3D0)
Se puede resolver de varias maneras, lo haré con la fórmula cuadrática
Esta dicta que:
![\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%5Cquad+x_%7B1%2C%5C%3A2%7D%3D%5Cfrac%7B-b%5Cpm+%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Nos da que este logaritomo es
![x=\frac{-\frac{\ln \left(3\right)}{\ln \left(10\right)}+\sqrt{\left(\frac{\ln \left(3\right)}{\ln \left(10\right)}\right)^2-4\left(-\frac{6\ln \left(3\right)}{\ln \left(10\right)}\right)\frac{\ln \left(3\right)}{\ln \left(10\right)}}}{2\frac{\ln \left(3\right)}{\ln \left(10\right)}}=2 x=\frac{-\frac{\ln \left(3\right)}{\ln \left(10\right)}+\sqrt{\left(\frac{\ln \left(3\right)}{\ln \left(10\right)}\right)^2-4\left(-\frac{6\ln \left(3\right)}{\ln \left(10\right)}\right)\frac{\ln \left(3\right)}{\ln \left(10\right)}}}{2\frac{\ln \left(3\right)}{\ln \left(10\right)}}=2](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%5Cfrac%7B%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%2810%5Cright%29%7D%2B%5Csqrt%7B%5Cleft%28%5Cfrac%7B%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%2810%5Cright%29%7D%5Cright%29%5E2-4%5Cleft%28-%5Cfrac%7B6%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%2810%5Cright%29%7D%5Cright%29%5Cfrac%7B%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%2810%5Cright%29%7D%7D%7D%7B2%5Cfrac%7B%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%2810%5Cright%29%7D%7D%3D2)
o
![x=\frac{-\frac{\ln \left(3\right)}{\ln \left(10\right)}-\sqrt{\left(\frac{\ln \left(3\right)}{\ln \left(10\right)}\right)^2-4\left(-\frac{6\ln \left(3\right)}{\ln \left(10\right)}\right)\frac{\ln \left(3\right)}{\ln \left(10\right)}}}{2\frac{\ln \left(3\right)}{\ln \left(10\right)}}=-3 x=\frac{-\frac{\ln \left(3\right)}{\ln \left(10\right)}-\sqrt{\left(\frac{\ln \left(3\right)}{\ln \left(10\right)}\right)^2-4\left(-\frac{6\ln \left(3\right)}{\ln \left(10\right)}\right)\frac{\ln \left(3\right)}{\ln \left(10\right)}}}{2\frac{\ln \left(3\right)}{\ln \left(10\right)}}=-3](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%5Cfrac%7B%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%2810%5Cright%29%7D-%5Csqrt%7B%5Cleft%28%5Cfrac%7B%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%2810%5Cright%29%7D%5Cright%29%5E2-4%5Cleft%28-%5Cfrac%7B6%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%2810%5Cright%29%7D%5Cright%29%5Cfrac%7B%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%2810%5Cright%29%7D%7D%7D%7B2%5Cfrac%7B%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%2810%5Cright%29%7D%7D%3D-3)
x=-3, x=2
Nos da una ecuación cuadrática
Se puede resolver de varias maneras, lo haré con la fórmula cuadrática
Esta dicta que:
Nos da que este logaritomo es
o
x=-3, x=2
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