How to solve the following equation

log₃ (3^x-8)=2-x

Thank you

Thanks for asking the question.Basically in this question we have to form an algebraic equation and then solve it to find x.log3(3^x -8) = 2-x.=> 3^x - 8 = 3^(2-x)=> 3^x -8 = 3^2/3^x=>3^x - 8 = 9/3^xLet, 3^x = y Therefore, y - 8 = 9/y.=> y^2 - 8y = 9On solving the above quadratic equation, we get y = 9 and y = -1=> 3^x = 9 and 3^x = -1.Since, 3^x = -1 is not possible.Therefore 3^x = 9=> x = 2.Hence, only solution is x=2