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Hola.
![1+\left(9^x\right)=\left(3^x\right)+1+\left(3^x-1\right) \\ \\ \mathrm{Convertir\:}9^x\mathrm{\:a\:base\:}3 = (3^2)^x \\ \\ 1+\left(3^2\right)^x=3^x+1+3^x-1 \\ \\ \mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \left(a^b\right)^c=a^{bc} \\ \\ 1+3^{2x}=3^x+1+3^x-1 \\ \\ \mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \:a^{bc}=\left(a^b\right)^c \\ \\ 1+\left(3^x\right)^2=3^x+1+3^x-1 \\ \\ 3^x=u \\ \\ Reescribimos: 1+\left(u\right)^2=u+1+u-1 \\ \\ 1+u^2=u+1+u-1 1+\left(9^x\right)=\left(3^x\right)+1+\left(3^x-1\right) \\ \\ \mathrm{Convertir\:}9^x\mathrm{\:a\:base\:}3 = (3^2)^x \\ \\ 1+\left(3^2\right)^x=3^x+1+3^x-1 \\ \\ \mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \left(a^b\right)^c=a^{bc} \\ \\ 1+3^{2x}=3^x+1+3^x-1 \\ \\ \mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \:a^{bc}=\left(a^b\right)^c \\ \\ 1+\left(3^x\right)^2=3^x+1+3^x-1 \\ \\ 3^x=u \\ \\ Reescribimos: 1+\left(u\right)^2=u+1+u-1 \\ \\ 1+u^2=u+1+u-1](https://tex.z-dn.net/?f=1%2B%5Cleft%289%5Ex%5Cright%29%3D%5Cleft%283%5Ex%5Cright%29%2B1%2B%5Cleft%283%5Ex-1%5Cright%29++%5C%5C++%5C%5C+%5Cmathrm%7BConvertir%5C%3A%7D9%5Ex%5Cmathrm%7B%5C%3Aa%5C%3Abase%5C%3A%7D3+%3D+%283%5E2%29%5Ex+%5C%5C++%5C%5C+1%2B%5Cleft%283%5E2%5Cright%29%5Ex%3D3%5Ex%2B1%2B3%5Ex-1++%5C%5C++%5C%5C+%5Cmathrm%7BAplicar%5C%3Alas%5C%3Aleyes%5C%3Ade%5C%3Alos%5C%3Aexponentes%7D%3A%5Cquad+%5Cleft%28a%5Eb%5Cright%29%5Ec%3Da%5E%7Bbc%7D++%5C%5C++%5C%5C+1%2B3%5E%7B2x%7D%3D3%5Ex%2B1%2B3%5Ex-1+%5C%5C++%5C%5C+%5Cmathrm%7BAplicar%5C%3Alas%5C%3Aleyes%5C%3Ade%5C%3Alos%5C%3Aexponentes%7D%3A%5Cquad+%5C%3Aa%5E%7Bbc%7D%3D%5Cleft%28a%5Eb%5Cright%29%5Ec++%5C%5C++%5C%5C+1%2B%5Cleft%283%5Ex%5Cright%29%5E2%3D3%5Ex%2B1%2B3%5Ex-1+%5C%5C++%5C%5C+3%5Ex%3Du++%5C%5C++%5C%5C+Reescribimos%3A+1%2B%5Cleft%28u%5Cright%29%5E2%3Du%2B1%2Bu-1+%5C%5C++%5C%5C+1%2Bu%5E2%3Du%2B1%2Bu-1++)
![u+u+1-1 = 2u \\ \\ 1+u^2=2u \\ \\ 1+u^2-2u=0 \\ \\ Formula: {x= \dfrac{-b\pm \sqrt{b ^{2}-4(a)(c) } }{2(a)} } \\ \\ a=1, b= -2, c=1, \\ \\ Reemplazamos: {x= \dfrac{-(-2)\pm\sqrt{(-2) ^{2}-4(1)(1) } }{2(1)} } \\ \\ Resultado: u=1 u+u+1-1 = 2u \\ \\ 1+u^2=2u \\ \\ 1+u^2-2u=0 \\ \\ Formula: {x= \dfrac{-b\pm \sqrt{b ^{2}-4(a)(c) } }{2(a)} } \\ \\ a=1, b= -2, c=1, \\ \\ Reemplazamos: {x= \dfrac{-(-2)\pm\sqrt{(-2) ^{2}-4(1)(1) } }{2(1)} } \\ \\ Resultado: u=1](https://tex.z-dn.net/?f=u%2Bu%2B1-1+%3D+2u+%5C%5C++%5C%5C+1%2Bu%5E2%3D2u++%5C%5C++%5C%5C+1%2Bu%5E2-2u%3D0++%5C%5C++%5C%5C+Formula%3A+%7Bx%3D+%5Cdfrac%7B-b%5Cpm+%5Csqrt%7Bb+%5E%7B2%7D-4%28a%29%28c%29+%7D+%7D%7B2%28a%29%7D+%7D++%5C%5C++%5C%5C+a%3D1%2C+b%3D+-2%2C+c%3D1%2C+%5C%5C++%5C%5C++Reemplazamos%3A+%7Bx%3D+%5Cdfrac%7B-%28-2%29%5Cpm%5Csqrt%7B%28-2%29+%5E%7B2%7D-4%281%29%281%29+%7D+%7D%7B2%281%29%7D+%7D+%5C%5C++%5C%5C+Resultado%3A+u%3D1)
Ahora..
![3^x=1 \\ \\ \ln \left(3^x\right)=\ln \left(1\right) \\ \\ \ln \left(1\right) = 0 \\ \\ x\ln \left(3\right)=0 \\ \\ \frac{x\ln \left(3\right)}{\ln \left(3\right)}=\frac{0}{\ln \left(3\right)} \\ \\ x=0 3^x=1 \\ \\ \ln \left(3^x\right)=\ln \left(1\right) \\ \\ \ln \left(1\right) = 0 \\ \\ x\ln \left(3\right)=0 \\ \\ \frac{x\ln \left(3\right)}{\ln \left(3\right)}=\frac{0}{\ln \left(3\right)} \\ \\ x=0](https://tex.z-dn.net/?f=3%5Ex%3D1++%5C%5C++%5C%5C+%5Cln+%5Cleft%283%5Ex%5Cright%29%3D%5Cln+%5Cleft%281%5Cright%29++%5C%5C++%5C%5C++%5Cln+%5Cleft%281%5Cright%29+%3D+0+%5C%5C++%5C%5C+x%5Cln+%5Cleft%283%5Cright%29%3D0++%5C%5C++%5C%5C+%5Cfrac%7Bx%5Cln+%5Cleft%283%5Cright%29%7D%7B%5Cln+%5Cleft%283%5Cright%29%7D%3D%5Cfrac%7B0%7D%7B%5Cln+%5Cleft%283%5Cright%29%7D++%5C%5C++%5C%5C+x%3D0)
¡Espero haberte ayudado, saludos...!
Ahora..
¡Espero haberte ayudado, saludos...!
kata42:
gracias
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