1) \( \left(\sqrt{3^{x+1}}-\frac{1}{9}\right)\left(4^{x}-7 \cdot 2^{x}+10\right)<0 \)

To tackle this inequality, let's begin by simplifying both factors of the product. First, we deal with the left part of the inequality: \[ \sqrt{3^{x+1}} = \sqrt{3 \cdot 3^x} = \sqrt{3} \cdot 3^{x/2}. \] Thus, the expression becomes: \[ \sqrt{3} \cdot 3^{x/2} - \frac{1}{9} < 0. \] Bringing \(\frac{1}{9}\) to the left side gives us: \[ \sqrt{3} \cdot 3^{x/2} < \frac{1}{9}. \] To facilitate solving this, we can multiply both sides by \(9\): \[ 9\sqrt{3} \cdot 3^{x/2} < 1. \] Rearranging this, we have: \[ 3^{x/2} < \frac{1}{9\sqrt{3}}. \] Since \(\frac{1}{9} = 3^{-2}\) and \(9\sqrt{3} = 3^{2} \cdot 3^{1/2} = 3^{2.5} = 3^{5/2}\), we have: \[ 3^{x/2} < 3^{-5/2}. \] Taking logarithms base \(3\): \[ \frac{x}{2} < -\frac{5}{2}, \] which simplifies to: \[ x < -5. \] Next, we analyze the right part of the inequality: \[ 4^{x} - 7 \cdot 2^{x} + 10 < 0. \] Recognizing \(4^x\) as \((2^2)^x = (2^x)^2\), let \(y = 2^{x}\). Then the expression becomes: \[ y^2 - 7y + 10 < 0. \] Factoring this gives: \[ (y - 5)(y - 2) < 0. \] The critical points are \(y = 2\) and \(y = 5\). We determine the intervals: - For \(y < 2\), both factors are negative, so the product is positive. - For \(2 < y < 5\), the product is negative. - For \(y > 5\), both factors are positive, thus the product is positive. This inequality holds when: \[ 2 < y < 5. \] Converting back to \(x\) using \(y = 2^x\): \[ 2 < 2^{x} < 5. \] Taking logarithms (base \(2\)): \[ 1 < x < \log_2(5). \] To visualize the solution set: 1. From the first factor, \(x < -5\). 2. From the second factor, \(1 < x < \log_2(5)\). These two parts do not overlap, which means no \(x\) satisfies both inequalities simultaneously. Thus, the complete solution of the original inequality \( \left(\sqrt{3^{x+1}}-\frac{1}{9}\right)\left(4^{x}-7 \cdot 2^{x}+10\right)<0 \) is: \[ \text{No solution.} \]