2.2 Solve for \( x \) 2.2.1 \( \quad \sqrt[x]{27}=2187 \) 2.2.2 \( \quad 4^{x}-16=6.2^{2} \) 2.3 Glven that \( x=\sqrt{3}-2 \), simplify \( \frac{x^{2}+1}{x^{2}-5} \) without using a calculator. (Give your answer in simplest surd form)

To solve \( \sqrt[x]{27} = 2187 \), we first rewrite it in its exponential form: \( 27^{\frac{1}{x}} = 2187 \). Noticing that \( 27 = 3^3 \) and \( 2187 = 3^7 \), we have \( (3^3)^{\frac{1}{x}} = 3^7 \). This simplifies to \( \frac{3}{x} = 7 \), leading us to \( x = \frac{3}{7} \). For \( 4^x - 16 = 6 \cdot 2^2 \) or \( 6 \cdot 4 \), we rewrite \( 16 \) as \( 4^2 \) and equate: \( 4^x - 4^2 = 24 \). Then, we have \( 4^x = 24 + 16 \) or \( 4^x = 40 \). Taking the logarithm base 4, we find \( x = \log_4(40) \). Now, let’s simplify \( \frac{x^2 + 1}{x^2 - 5} \) with \( x = \sqrt{3} - 2 \). First, calculate \( x^2 \): \[ x^2 = (\sqrt{3} - 2)^2 = 3 - 4\sqrt{3} + 4 = 7 - 4\sqrt{3} \] Next, substitute \( x^2 \) back into the expression: \[ x^2 + 1 = (7 - 4\sqrt{3}) + 1 = 8 - 4\sqrt{3} \] \[ x^2 - 5 = (7 - 4\sqrt{3}) - 5 = 2 - 4\sqrt{3} \] Now, we simplify \( \frac{8 - 4\sqrt{3}}{2 - 4\sqrt{3}} \): Factoring out 4 yields: \[ = \frac{4(2 - \sqrt{3})}{2(1 - 2\sqrt{3})} = \frac{2 - \sqrt{3}}{1 - 2\sqrt{3}} \] To rationalize the denominator: \[ = \frac{(2 - \sqrt{3})(1 + 2\sqrt{3})}{(1 - 2\sqrt{3})(1 + 2\sqrt{3})} \] The denominator simplifies to \( 1 - 12 = -11 \). Calculate the numerator: \[ 2 + 4\sqrt{3} - \sqrt{3}-6 = -4 + 3\sqrt{3} \] Thus the final expression, simplified in surd form, is: \[ \frac{3\sqrt{3} - 4}{-11} = \frac{4 - 3\sqrt{3}}{11} \] So, the final results for the problems are: 1. \( x = \frac{3}{7} \) 2. \( x = \log_4(40) \) 3. \( \frac{4 - 3\sqrt{3}}{11} \)