Respuestas
Respuesta dada por:
2
Log3(4x-5)-log3(x+1)=0
Resolvemos:
![Log3(4x-5)-log3(x+1)=0 \\ \\ \log _{10}\left(3\right)\cdot 4x-\log _{10}\left(3\right)\cdot \:5-\log _{10}\left(3\right)x-\log _{10}\left(3\right) =0 \\ \\ \textbf{Simplificamos:} \\ \\ 4\log _{10}\left(3\right)x-5\log _{10}\left(3\right)-\log _{10}\left(3\right)x-\log _{10}\left(3\right)=0 \\ \\ 4\log _{10}\left(3\right)x-6\log _{10}\left(3\right)-\log _{10}\left(3\right)x=0 Log3(4x-5)-log3(x+1)=0 \\ \\ \log _{10}\left(3\right)\cdot 4x-\log _{10}\left(3\right)\cdot \:5-\log _{10}\left(3\right)x-\log _{10}\left(3\right) =0 \\ \\ \textbf{Simplificamos:} \\ \\ 4\log _{10}\left(3\right)x-5\log _{10}\left(3\right)-\log _{10}\left(3\right)x-\log _{10}\left(3\right)=0 \\ \\ 4\log _{10}\left(3\right)x-6\log _{10}\left(3\right)-\log _{10}\left(3\right)x=0](https://tex.z-dn.net/?f=Log3%284x-5%29-log3%28x%2B1%29%3D0+%5C%5C++%5C%5C+%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%5Ccdot+4x-%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%5Ccdot+%5C%3A5-%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x-%5Clog+_%7B10%7D%5Cleft%283%5Cright%29+%3D0+%5C%5C+%5C%5C+%5Ctextbf%7BSimplificamos%3A%7D+%5C%5C+%5C%5C+4%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x-5%5Clog+_%7B10%7D%5Cleft%283%5Cright%29-%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x-%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%3D0+%5C%5C+%5C%5C+4%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x-6%5Clog+_%7B10%7D%5Cleft%283%5Cright%29-%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x%3D0+)
![3\log _{10}\left(3\right)x-6\log _{10}\left(3\right)=0 \\ \\ 3\log _{10}\left(3\right)x-6\log _{10}\left(3\right)+6\log _{10}\left(3\right)=0+6\log _{10}\left(3\right) \\ \\ 3\log _{10}\left(3\right)x=6\log _{10}\left(3\right) \\ \\ \dfrac{3\log _{10}\left(3\right)x}{3\log _{10}\left(3\right)}=\dfrac{6\log _{10}\left(3\right)}{3\log _{10}\left(3\right)} \\ \\ \boxed{\boxed{x=2}} \ \ $Respuesta$ 3\log _{10}\left(3\right)x-6\log _{10}\left(3\right)=0 \\ \\ 3\log _{10}\left(3\right)x-6\log _{10}\left(3\right)+6\log _{10}\left(3\right)=0+6\log _{10}\left(3\right) \\ \\ 3\log _{10}\left(3\right)x=6\log _{10}\left(3\right) \\ \\ \dfrac{3\log _{10}\left(3\right)x}{3\log _{10}\left(3\right)}=\dfrac{6\log _{10}\left(3\right)}{3\log _{10}\left(3\right)} \\ \\ \boxed{\boxed{x=2}} \ \ $Respuesta$](https://tex.z-dn.net/?f=3%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x-6%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%3D0++%5C%5C+%5C%5C+3%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x-6%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%2B6%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%3D0%2B6%5Clog+_%7B10%7D%5Cleft%283%5Cright%29+%5C%5C+%5C%5C+3%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x%3D6%5Clog+_%7B10%7D%5Cleft%283%5Cright%29+%5C%5C+%5C%5C+%5Cdfrac%7B3%5Clog+_%7B10%7D%5Cleft%283%5Cright%29x%7D%7B3%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%7D%3D%5Cdfrac%7B6%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%7D%7B3%5Clog+_%7B10%7D%5Cleft%283%5Cright%29%7D+%5C%5C+%5C%5C+%5Cboxed%7B%5Cboxed%7Bx%3D2%7D%7D+%5C+%5C+%24Respuesta%24)
¡Espero haberte ayudado, saludos... G.G.H!
Resolvemos:
¡Espero haberte ayudado, saludos... G.G.H!
Preguntas similares
hace 6 años
hace 6 años
hace 6 años
hace 9 años
hace 9 años
hace 9 años
hace 9 años
hace 9 años