\( \begin{array}{ll}\text { (c) } & -2^{x}=-16 \\ \text { (a) } & -2^{x}=-1 \\ \text { (a) } & 7^{x}=1 \\ \text { (a) } & 10^{10 x}=100 \\ \text { (a) } & 25^{x}=125 \\ \text { (i) } & 9^{3 x-1}=27\end{array} \)

1. For the equation \( -2^x = -16 \): - Interpret the left-hand side as \(-2^x = -(2^x)\). - Thus, we have \[ -(2^x) = -16. \] - Multiply both sides by \(-1\): \[ 2^x = 16. \] - Recognize that \(16 = 2^4\), so: \[ 2^x = 2^4. \] - Therefore, \( x = 4 \). 2. For the equation \( -2^x = -1 \): - Rewrite the equation as \[ -(2^x) = -1. \] - Multiply both sides by \(-1\): \[ 2^x = 1. \] - Since \(1 = 2^0\), we rewrite as: \[ 2^x = 2^0. \] - Thus, \( x = 0 \). 3. For the equation \( 7^x = 1 \): - Note that \(1 = 7^0\), so we have: \[ 7^x = 7^0. \] - Therefore, \( x = 0 \). 4. For the equation \( 10^{10x} = 100 \): - Recall that \(100 = 10^2\). So, rewrite the equation as: \[ 10^{10x} = 10^2. \] - Equate the exponents: \[ 10x = 2. \] - Solve for \( x \): \[ x = \frac{2}{10} = \frac{1}{5}. \] 5. For the equation \( 25^x = 125 \): - Express the bases as powers of \(5\): \(25 = 5^2\) and \(125 = 5^3\). - The equation becomes: \[ (5^2)^x = 5^3. \] - Simplify: \[ 5^{2x} = 5^3. \] - Equate the exponents: \[ 2x = 3. \] - Solve for \( x \): \[ x = \frac{3}{2}. \] 6. For the equation \( 9^{3x-1} = 27 \): - Write the numbers in base \(3\): \(9 = 3^2\) and \(27 = 3^3\). - The equation becomes: \[ (3^2)^{3x-1} = 3^3. \] - Simplify using the power rule: \[ 3^{2(3x-1)} = 3^3. \] - Equate the exponents: \[ 2(3x-1) = 3. \] - Distribute the \(2\): \[ 6x - 2 = 3. \] - Solve for \( x \) by adding \(2\) to both sides: \[ 6x = 5. \] - Thus, \[ x = \frac{5}{6}. \] Final answers: 1. \( x = 4 \) 2. \( x = 0 \) 3. \( x = 0 \) 4. \( x = \frac{1}{5} \) 5. \( x = \frac{3}{2} \) 6. \( x = \frac{5}{6} \)