How would I solve this equation?

Question

I know that since there are exponents they have to have the same base but I can't figure it out. Thanks for any help!

 

9^8x · (1/3)^ (4x + 2) = 27 · (3^x)^−4

Answer 1

 

Hi Sky,

 

Looking at the expression you typed in,

 

I think the original problem is

 

     9^8x(1/3)^4x+2 = 27(3^x)^-4     (please, correct me if this is not the case)

 

Let's make all items have the same base of 3.

 

First, 9^8x = (3²)^8x

 

              = 3^2·8x

 

              = 3^16x

 

      (1/3)^4x+2 = (3^-1)^4x+2

 

                    = 3^-1·(4x+2)

 

                    = 3^-4x-2

 

      27 = 3³

 

      (3^x)^-4 = 3^x·(-4)

 

               = 3^-4x

 

Putting everything into the original equation,

 

     9^8x(1/3)^4x+2 = 27(3^x)^-4

 

     3^16x3^-4x-2 = 3³3^-4x

 

     3^16x-4x-2 = 3^3-4x

 

     3^12x-2 = 3^-4x+3

 

Now, we get

 

     12x - 2 = -4x + 3

 

Solving the above linear equation,

 

we get x = 5/16.

 

So, the answer is x = 5/16.

 

Hope this helps.

 

 

 

 

Answer 2

9^8x(1/3)^4x+2 = 27(3^x)^-4

 

Express both sides of the equation as a power of 3:

 

(3²)^8x(3^-1)^4x+2 =(3)³(3^x)^-4

 

3^16x3^-4x-2 = 3³3^-4x

 

3^12x-2 = 3^3-4x

 

12x-2 = 3-4x

 

16x = 5

 

   x = 5/16