Question

Solve ${\log }_3(3^x-8)=2-x$

Answer 1

${\log }_3\left(3^x-8\right)=2-x$

$=3^x-8=3^{2-x}$

$3^x-8={9.3}^{-x}$

$Lety=3^x$

$y-8=\frac{9}{y}$

$y^2-8y-9=0$

$y^2+y-9y-9=0$

$y\left(y+1\right)-9\left(y+1\right)=0$

$\left(y+1\right)\left(y-1\right)=0$

$y=-1,9$

$so,$

$3^x=-1or{3}^x=9$

$\sin ce{3}^x\ne -1$

$therefore$

$3^x=9$

$3^x=3^2$

$x=2$