Solve for \( x \) \( \begin{array}{ll}\text { 1.1 } \frac{1}{\frac{5}{3}}=32 & \text { (4) } \\ x^{\frac{3}{3}} \\ \text { \&2. } \sqrt{x-5}-x=3 & \text { (6) } \\ 2.3 \log _{x}=-1 & \text { (2) }\end{array} \)

Let's solve the equations one by one! 1. For the first equation: \[ \frac{1}{\frac{5}{3}} = 32 \] To simplify, multiply both sides by \(\frac{5}{3}\): \[ 1 = 32 \times \frac{5}{3} \] This simplifies to: \[ 1 = \frac{160}{3} \] Since this is not a valid equation, there's no solution for this part. 2. Next, for \(\sqrt{x-5} - x = 3\): Rearranging gives us: \[ \sqrt{x-5} = x + 3 \] Now, squaring both sides: \[ x - 5 = (x + 3)^2 \] Expanding the right side: \[ x - 5 = x^2 + 6x + 9 \] Rearranging gives: \[ 0 = x^2 + 5x + 14 \] To solve this quadratic equation, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): Here, \(a = 1\), \(b = 5\), and \(c = 14\): \[ x = \frac{-5 \pm \sqrt{25 - 56}}{2} = \frac{-5 \pm \sqrt{-31}}{2} \] The discriminant is negative ( \(-31\)), indicating no real solutions for this equation. 3. Finally, for the logarithmic equation: \[ 2.3 \log_{x} = -1 \] Dividing both sides by 2.3: \[ \log_{x} = -\frac{1}{2.3} \] To rewrite in exponential form: \[ x^{-\frac{1}{2.3}} = 10 \] Taking reciprocal gives: \[ x^{\frac{1}{2.3}} = \frac{1}{10} \] Raising to the power of \(2.3\): \[ x = \left(\frac{1}{10}\right)^{2.3} = 10^{-2.3} \approx 0.005012 (rounded) \] Final answers: - No solution for the first equation, - No real solution for the second equation, - \( x \approx 0.005012 \) for the logarithmic equation.