Question

3^x=6,calcula g^x-1

Answer 1

Nos dan:

$3^x=6$

Si f(x) = g(x), entonces In( f(x) ) = In( g(x) ):

$\ln \left(3^x\right)=\ln \left(6\right)$

Aplicamos propiedades de los logaritmos:

$\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)\\\\\ln \left(3^x\right)=x\ln \left(3\right)\\\\x\ln \left(3\right)=\ln \left(6\right)\\\\\frac{x\ln \left(3\right)}{\ln \left(3\right)}=\frac{\ln \left(6\right)}{\ln \left(3\right)}\\\\x=\frac{\ln \left(6\right)}{\ln \left(3\right)}$

Ahora, resolvemos lo otro:

$g^{\frac{\ln \left(6\right)}{\ln \left(3\right)}-1}\\\\=\frac{\ln \left(6\right)}{\ln \left(3\right)}-\frac{1\cdot \ln \left(3\right)}{\ln \left(3\right)}\\\\=\frac{\ln \left(6\right)-1\cdot \ln \left(3\right)}{\ln \left(3\right)}\\\\=\frac{\ln \left(6\right)-\ln \left(3\right)}{\ln \left(3\right)}\\\\=\frac{\ln \left(2\right)}{\ln \left(3\right)}\\\\=g^{\frac{\ln \left(2\right)}{\ln \left(3\right)}}$

Espero haberte ayudado