a. \( 10^{-\sqrt{x}}=100^{1-x} \)

To solve the equation \( 10^{-\sqrt{x}} = 100^{1-x} \), we can start by rewriting \( 100 \) in terms of base \( 10 \): \[ 100 = 10^2 \] Thus, we can rewrite \( 100^{1-x} \) as: \[ 100^{1-x} = (10^2)^{1-x} = 10^{2(1-x)} = 10^{2 - 2x} \] Now, we can rewrite the original equation: \[ 10^{-\sqrt{x}} = 10^{2 - 2x} \] Since the bases are the same, we can set the exponents equal to each other: \[ -\sqrt{x} = 2 - 2x \] Next, we can rearrange this equation to isolate the square root: \[ \sqrt{x} = 2x - 2 \] Now, we will square both sides to eliminate the square root: \[ x = (2x - 2)^2 \] Expanding the right side: \[ x = 4x^2 - 8x + 4 \] Rearranging the equation gives us: \[ 0 = 4x^2 - 9x + 4 \] Now, we can use the quadratic formula to solve for \( x \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 4 \), \( b = -9 \), and \( c = 4 \): \[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4} \] \[ x = \frac{9 \pm \sqrt{81 - 64}}{8} \] \[ x = \frac{9 \pm \sqrt{17}}{8} \] Thus, the solutions for \( x \) are: \[ x = \frac{9 + \sqrt{17}}{8} \quad \text{and} \quad x = \frac{9 - \sqrt{17}}{8} \] Now, we need to check if both solutions are valid in the context of the original equation. Since \( \sqrt{x} \) must be non-negative, we need \( x \geq 0 \). Calculating the approximate values: 1. For \( x = \frac{9 + \sqrt{17}}{8} \): - \( \sqrt{17} \approx 4.123 \) - \( x \approx \frac{9 + 4.123}{8} \approx \frac{13.123}{8} \approx 1.640 \) (valid) 2. For \( x = \frac{9 - \sqrt{17}}{8} \): - \( x \approx \frac{9 - 4.123}{8} \approx \frac{4.877}{8} \approx 0.6096 \) (valid) Both solutions are valid. Therefore, the final solutions are: \[ \boxed{\frac{9 + \sqrt{17}}{8}} \quad \text{and} \quad \boxed{\frac{9 - \sqrt{17}}{8}} \]