Question

Solve the equation 3^(2x) -4*3^(x+1) + 27 = 0. use rules of exponents to change 3^(x+1)​

Answer 1

Answer:

$\huge\boxed{x=1\ \vee\ x=2}$

Step-by-step explanation:

$3^{2x}-4\cdot3^{x+1}+27=0\\\\\text{use}\ (a^n)^m=a^{nm}\ \text{and}\ a^n\cdot a^m=a^{n+m}\\\\\left(3^x\right)^2-4\cdot3^x\cdot3^1+27=0\\\\\left(3^x\right)^2-12\cdot3^x+27=0\\\\\text{substitute}\ 3^x=t>0\\\\t^2-12t+27=0\\\\t^2-3t-9t+27=0\\\\t(t-3)-9(t-3)=0\\\\(t-3)(t-9)+0\iff t-3=0\ \vee\ t-9=0\\\\t-3=0\qquad\text{add 3 to both sides}\\\boxed{t=3}\\\\t-9=0\qquad\text{add 9 to both sides}\\\boxed{t=9}$

$\text{We return to substitution:}\\\\3^x=t\\\\3^x=3\ \vee\ 3^x=9\\\\3^x=3^1\ \vee\ 3^x=3^2\\\\\boxed{x=1}\ \vee\ \boxed{x=2}$