Question

What is the solution to this equation? (1/27)2-x =9^3x

Answer 1

Answer:

 x = -2

Step-by-step explanation:

Prime factorize, 27 and 9

27 = 3 *3 * 3 = 3³

  9 = 3*3 = 3²

$\sf \left(\dfrac{1}{27}\right)^{2-x}=9^{3x}$

 $\left(\dfrac{1}{3^3}\right)^{2-x}=9^{3x}\\\\(3^{-3})^{2-x}=(3^2)^{3x}\\\\3^{(-3)*(2-x)}=3^{2*3x}\\\\3^{-6+3x} = 3^{6x}$

Bases are same, so now compare the exponents

     -6 + 3x = 6x

            3x = 6x + 6

       3x - 6x = 6

             -3x = 6

               x  = 6/(-3)

               $\sf \boxed{\bf x = -2}$

Exponent law:

         $(x^m)^n=x^{m*n}\\\\\left(\dfrac{1}{a^{m}}\right)=a^{-m}$