To solve the equation \( 27^x = \frac{3}{9^{x-3}} \), we first express all numbers as powers of 3.
We know:
- \( 27 = 3^3 \)
- \( 9 = 3^2 \)
Thus, we can rewrite both sides:
\[ 27^x = (3^3)^x = 3^{3x} \]
For the right-hand side:
\[ 9^{x-3} = (3^2)^{x-3} = 3^{2(x-3)} = 3^{2x - 6} \]
So the equation becomes:
\[ 3^{3x} = \frac{3}{3^{2x - 6}} \]
Next, we can simplify the right-hand side:
\[ \frac{3}{3^{2x - 6}} = 3^{1 - (2x - 6)} = 3^{1 - 2x + 6} = 3^{7 - 2x} \]
Now we have:
\[ 3^{3x} = 3^{7 - 2x} \]
Since the bases are the same, we can set the exponents equal to each other:
\[ 3x = 7 - 2x \]
Now, we will solve for \( x \). Adding \( 2x \) to both sides gives:
\[ 3x + 2x = 7 \]
\[ 5x = 7 \]
Now, divide both sides by 5:
\[ x = \frac{7}{5} \]
Thus, the solution to the equation is:
\[ \boxed{\frac{7}{5}} \]