\begin{tabular}{|l}\hline Solve the following equations: \\ \hline (a) \( 2^{x}=\frac{1}{8} \) \\ \hline (d) \( 3^{x}=\frac{1}{243} \) \\ \hline (g) \( 2^{x}=0,5 \) \\ \hline (j) \( \left(\frac{1}{4}\right)^{x}=32 \) \\ \hline (m) \( 3.3^{x}=81 \) \\ \hline (p) \( 5.4^{x-2}=80 \) \\ \hline\end{tabular}

To solve the equations, we can use the properties of exponents and logarithms. For (a) \( 2^{x}=\frac{1}{8} \): Since \( \frac{1}{8} = 2^{-3} \), we have \( 2^{x} = 2^{-3} \). Therefore, \( x = -3 \). For (d) \( 3^{x}=\frac{1}{243} \): Knowing that \( \frac{1}{243} = 3^{-5} \), we get \( 3^{x} = 3^{-5} \), leading to \( x = -5 \). For (g) \( 2^{x}=0.5 \): Recognizing that \( 0.5 = \frac{1}{2} = 2^{-1} \), it follows that \( x = -1 \). For (j) \( \left(\frac{1}{4}\right)^{x}=32 \): Since \( \frac{1}{4} = 2^{-2} \), we rewrite it as \( (2^{-2})^{x} = 2^{5} \). So, \( 2^{-2x} = 2^{5} \), which gives \( -2x = 5 \) or \( x = -\frac{5}{2} \). For (m) \( 3.3^{x}=81 \): First express \( 81 \) as \( 3^4 \), resulting in \( 3 \cdot 3^{x} = 3^{4} \), hence \( 3^{x+1} = 3^{4} \). Therefore, \( x + 1 = 4 \) leading to \( x = 3 \). For (p) \( 5 \cdot 4^{x-2}=80 \): Divide by 5 to get \( 4^{x-2} = 16 \). Since \( 16 = 4^2 \), we have \( 4^{x-2} = 4^{2} \), so \( x - 2 = 2 \), thus \( x = 4 \). So, the solutions are: (a) \( x = -3 \) (d) \( x = -5 \) (g) \( x = -1 \) (j) \( x = -\frac{5}{2} \) (m) \( x = 3 \) (p) \( x = 4 \)