Solve for \( x \) : \( \begin{array}{ll}2.2 .1 & 2-16 x^{-\frac{1}{3}}=0 \\ 2.2 .2 & 4^{x}+8=9.2^{x} \\ 2.2 .3 & \sqrt[x]{9}=243\end{array} \) Simplify fully:
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Let’s tackle each part of the equation step by step. ### 2.2.1: \( 2 - 16x^{-\frac{1}{3}} = 0 \) First, rearranging the equation gives: \[ 16x^{-\frac{1}{3}} = 2 \] Dividing both sides by 16: \[ x^{-\frac{1}{3}} = \frac{2}{16} = \frac{1}{8} \] Taking the reciprocal: \[ x^{\frac{1}{3}} = 8 \] Now, cubing both sides to solve for \( x \): \[ x = 8^3 = 512 \] ### 2.2.2: \( 4^x + 8 = 9 \cdot 2^x \) Rewrite \( 4^x \) as \( (2^2)^x = 2^{2x} \): \[ 2^{2x} + 8 = 9 \cdot 2^x \] Now, let \( y = 2^x \), converting the equation to: \[ y^2 + 8 = 9y \] Rearranging gives: \[ y^2 - 9y + 8 = 0 \] Factoring: \[ (y - 1)(y - 8) = 0 \] So, \( y = 1 \) or \( y = 8 \). Since \( y = 2^x \): - If \( 2^x = 1 \), then \( x = 0 \). - If \( 2^x = 8 \), then \( 2^x = 2^3 \), so \( x = 3 \). Thus, the solutions are \( x = 0 \) and \( x = 3 \). ### 2.2.3: \( \sqrt[x]{9} = 243 \) Rewrite the equation: \[ 9^{\frac{1}{x}} = 243 \] Recognizing \( 9 = 3^2 \) and \( 243 = 3^5 \): \[ (3^2)^{\frac{1}{x}} = 3^5 \] This simplifies to: \[ 3^{\frac{2}{x}} = 3^5 \] Setting the exponents equal to each other: \[ \frac{2}{x} = 5 \] Cross-multiplying to solve for \( x \): \[ 2 = 5x \implies x = \frac{2}{5} \] ### Summary of Solutions: - **2.2.1:** \( x = 512 \) - **2.2.2:** \( x = 0 \) and \( x = 3 \) - **2.2.3:** \( x = \frac{2}{5} \)
