\( -x=\frac{21}{3^{x}} \) (c) \( \frac{6^{x} \cdot 9^{x+1} \cdot 2}{27^{x+1} \cdot 2^{x-1}} \) (f) \( \frac{9^{x}+3^{2 x+1}}{18^{x} \cdot 2^{1-x}} \) (i) \( \frac{2 \cdot 3^{x+2}+3^{x-3}}{5 \cdot 3^{x-2}} \)

To solve the equation \( -x=\frac{21}{3^{x}} \), we can start by isolating \( 3^{x} \). Rearranging gives us \( 3^{x} = -\frac{21}{x} \). As \( x \) approaches negative infinity, the right side approaches 0, but since \( 3^{x} \) is strictly positive for all real \( x \), we need to look only at \( x < 0 \) for potential solutions and analyze it graphically for quicker insights. Now, for the expression \( \frac{6^{x} \cdot 9^{x+1} \cdot 2}{27^{x+1} \cdot 2^{x-1}} \), we can simplify this by expressing each base in terms of powers of 3 and 2. Notice that \( 6=2\cdot3\), \( 9=3^{2} \), and \( 27=3^{3} \). By plugging in these expressions and simplifying, we can combine similar terms to have a cleaner result that highlights how exponents add when bases are the same. This will lead you to a final simplified expression that you can interpret easily. For the equation \( \frac{9^{x}+3^{2 x+1}}{18^{x} \cdot 2^{1-x}} \), take note that \( 9^{x} = (3^{2})^{x} = 3^{2x} \) and \( 18 = 2 \cdot 9 \), which will help you express everything in terms of a uniform base for easier handling. The numerator can be combined effectively once you rewrite \( 3^{2x+1} \) as \( 3^{2x} \cdot 3^{1} \) and consolidate the whole expression under a common theme—this often opens up cancellation opportunities. Lastly, for \( \frac{2 \cdot 3^{x+2}+3^{x-3}}{5 \cdot 3^{x-2}} \), factor out \( 3^{x} \) from the numerator to harness the power of simplification. This lets you express everything in a more digestible way—much to the relief of your mathematical endeavors! Remember, simplifying before diving into calculations can save time and help avoid common pitfalls.