Solve for \( x \) : 2. \( \begin{array}{ll}2.2 .1 & \left(\frac{1}{2}\right)^{x}=32 \\ 2.2 .2 & \sqrt[3]{\frac{1}{x^{2}}}=4 \\ 2.2 .3 & 2^{x}-\frac{12}{2^{x}}=-4 \\ \text { WITHOUT using a calculator, show that } \frac{\sqrt{2}}{\sqrt{2}+1}+\frac{4}{\sqrt{2}} \text { simplifies to } 2+\sqrt{2} .\end{array} \)

Let's solve each of these equations step by step! ### 2.2.1 To solve the equation \(\left(\frac{1}{2}\right)^{x}=32\): First, rewrite \(32\) as a power of \(2\): \[ 32 = 2^5 \] So, we can express \(\frac{1}{2}\) as \(2^{-1}\): \[ (2^{-1})^x = 2^5 \] This simplifies to: \[ 2^{-x} = 2^5 \] Since the bases are the same, we can set the exponents equal to each other: \[ -x = 5 \implies x = -5 \] ### 2.2.2 To solve the equation \(\sqrt[3]{\frac{1}{x^{2}}}=4\): Cubing both sides gives: \[ \frac{1}{x^{2}} = 4^3 \] Calculating \(4^3\) gives: \[ \frac{1}{x^{2}} = 64 \] Now, taking the reciprocal gives: \[ x^2 = \frac{1}{64} \] Taking the square root of both sides yields: \[ x = \pm \frac{1}{8} \] ### 2.2.3 To solve the equation \(2^{x}-\frac{12}{2^{x}}=-4\): Let's set \(y = 2^{x}\). Therefore, the equation becomes: \[ y - \frac{12}{y} = -4 \] Multiplying through by \(y\) (assuming \(y \neq 0\)) gives: \[ y^2 + 4y - 12 = 0 \] Now we can factor this quadratic: \[ (y + 6)(y - 2) = 0 \] Thus, we have two solutions: \[ y + 6 = 0 \implies y = -6 \quad \text{(not applicable since } y = 2^x > 0\text{)} \] \[ y - 2 = 0 \implies y = 2 \implies 2^{x} = 2 \implies x = 1 \] ### Show that \(\frac{\sqrt{2}}{\sqrt{2}+1}+\frac{4}{\sqrt{2}}\) simplifies to \(2+\sqrt{2}\). First, let’s simplify \(\frac{4}{\sqrt{2}}\): \[ \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \] Now we have: \[ \frac{\sqrt{2}}{\sqrt{2}+1} + 2\sqrt{2} \] Let's find a common denominator for the first term: \[ \frac{\sqrt{2}}{\sqrt{2}+1} + \frac{2\sqrt{2}(\sqrt{2}+1)}{\sqrt{2}+1} = \frac{\sqrt{2} + 2\sqrt{2}(\sqrt{2}+1)}{\sqrt{2}+1} \] Expanding the numerator: \[ \sqrt{2} + 2\sqrt{2}(\sqrt{2}) + 2\sqrt{2} = \sqrt{2} + 4 + 2\sqrt{2} = 4 + 3\sqrt{2} \] So, we have: \[ \frac{4 + 3\sqrt{2}}{\sqrt{2}+1} \] Now we must rationalize the denominator by multiplying the numerator and denominator by \((\sqrt{2}-1)\): \[ \frac{(4 + 3\sqrt{2})(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{(4\sqrt{2} - 4 + 6 - 3\sqrt{2})}{(2-1)} = \frac{(4\sqrt{2} - 3\sqrt{2} + 2)}{1} = \sqrt{2} + 2 \] Therefore, we have shown: \[ \frac{\sqrt{2}}{\sqrt{2}+1} + \frac{4}{\sqrt{2}} = 2 + \sqrt{2} \] ### Final answers: - \(x = -5\) - \(x = \pm \frac{1}{8}\) - \(x = 1\) -