We have to solve: 9^(x-1) =27*3^x

9^(x-1) =27*3^x

convert all the terms to a power of 3

=> 3^2^(x - 1) = 3^3*3^x

=> 3^(2x - 2) = 3^(3 + x)

as the base 3 is the same, equate the exponent

=> 2x - 2 = 3 + x

=> x = 5

**The required solution of the equation is x = 5**

The first important step is to create matching bases both sides.

Since 9 is a power of 3, we'll write it as:

9 = 3^2

Also 27 is a power of 3:

27 = 3^3

Now, we'll apply the multiplication rule of 2 exponentials that have matching bases:

3^3*3^x = 3^(3 + x)

We'll re-write the equation:

3^2(x-1) = 3^(3 + x)

Since the bases are matching, we'll apply one to one rule:

2(x-1) = (3 + x)

We'll open the brackets:

2x - 2 = x + 3

We'll move all terms to one side:

2x - x - 2 - 3 = 0

We'll combine like terms:

x - 5 = 0

We'll add 5:

x = 5

**The requested solution of the given equation is x = 5.**