\( \left. \begin{array} { l } { 10 } \end{array} \right. \left. \begin{array} { l } { \frac { 5 ^ { x } - 1 } { 2 ^ { x } } \geq 0 } \\ \hline 11 \end{array} \right. \frac { 5 ^ { x } - 25 } { 3 ^ { x } + 9 } < 0 \)

To tackle this inequality problem, let’s first break down the two parts. The first inequality, \( \frac{5^x - 1}{2^x} \geq 0 \), indicates that \(5^x - 1\) must be non-negative, meaning \(5^x \geq 1\) or \(x \geq 0\). For the second part, \( \frac{5^x - 25}{3^x + 9} < 0 \), we need to establish when the numerator and denominator yield opposite signs. The numerator \( 5^x - 25 < 0 \) means \( 5^x < 25 \), or \( x < 2 \). The denominator \(3^x + 9\) is always positive for all real \(x\). Now, bringing it all together, we find that \(x\) must satisfy \(0 \leq x < 2\). So the solution set for \(x\) is \( [0, 2) \). Would you like to delve more into this topic? It can get quite fascinating!