Solve for \( x \) : (a) \( 3^{x}=1 \) (d) \( \left(\frac{1}{9}\right)^{x}=27 \) (g) \( 4\left(\frac{1}{4}\right)^{x-1}=8 \) (j) \( \frac{3^{2 x-1}}{3^{x}}=3 \) (m) \( 8^{-x}-2.4^{x-1}=0 \)

To solve each of these equations, we can use the properties of exponents and logarithms. Let’s delve into the solutions: (a) \( 3^{x} = 1 \) Since any base raised to the power of 0 equals 1, we know that \( x = 0 \). (d) \( \left(\frac{1}{9}\right)^{x} = 27 \) Rewriting \( \frac{1}{9} \) as \( 9^{-1} \), we have \( (9^{-1})^{x} = 27 \). Since \( 27 = 3^3 \) and \( 9 = 3^2 \), we get \( (3^{-2})^{x} = 3^3 \). Thus, \( -2x = 3 \) or \( x = -\frac{3}{2} \). (g) \( 4\left(\frac{1}{4}\right)^{x-1} = 8 \) This can be rewritten as \( 4 \cdot 4^{-(x-1)} = 8 \), simplifying to \( 4^{1 - (x - 1)} = 8 \) or \( 4^{2 - x} = 2^3 \). Converting bases gives us \( 4^{2 - x} = (2^2)^{2-x} = 2^{2(2-x)} = 2^3 \), leading to \( 2(2-x) = 3 \), from which \( 2 - x = \frac{3}{2} \) and \( x = -\frac{1}{2} \). (j) \( \frac{3^{2x-1}}{3^{x}} = 3 \) We can simplify this to \( 3^{2x - 1 - x} = 3 \) or \( 3^{x - 1} = 3^1 \). Therefore, \( x - 1 = 1 \), giving us \( x = 2 \). (m) \( 8^{-x} - 2 \cdot 4^{x - 1} = 0 \) Rewriting as \( 8^{-x} = 2 \cdot 4^{x - 1} \), and recognizing that \( 8 = 2^3 \) and \( 4 = 2^2 \), we can express that as \( (2^3)^{-x} = 2 \cdot (2^2)^{x-1} \). This leads to \( 2^{-3x} = 2^{1 + 2(x-1)} \), simplifying to \( -3x = 1 + 2x - 2 \), or \( -3x = 2x - 1 \). Therefore, \( -5x = -1 \) gives \( x = \frac{1}{5} \). To recap: - (a) \( x = 0 \) - (d) \( x = -\frac{3}{2} \) - (g) \( x = -\frac{1}{2} \) - (j) \( x = 2 \) - (m) \( x = \frac{1}{5} \)