To solve the inequality \(3^x + 2^{2x} \ge 5^x\), we need to explore when the left side of the inequality is greater than or equal to the right side as \(x\) varies.
- \(3^x + 2^{2x} \ge 5^x\) implies comparing the growth rates of the functions \(3^x\), \(2^{2x}\), and \(5^x\).
- Rewriting the expression:
- \(2^{2x} = (2^2)^x = 4^x\)
- Consider small values of \(x\) for analysis:
- For \(x = 0: 3^0 + 4^0 = 2 \cdot 1 = 1\) and \(5^0 = 1\). Thus, \(1 \ge 1\) holds true.
- For \(x = 1: 3^1 + 4^1 = 3 + 4 = 7\) and \(5^1 = 5\). Thus, \(7 \ge 5\) holds true.
- For \(x = 2: 3^2 + 4^2 = 9 + 16 = 25\) and \(5^2 = 25\). Thus, \(25 \ge 25\) holds true.
- For \(x = 3: 3^3 + 4^3 = 27 + 64 = 91\) and \(5^3 = 125\). Thus, \(91 \not\geq 125\) does not hold.
- Thus, \(x > 2\) makes \(5^x\) grow faster than the sum \(3^x + 4^x,\) indicating \(x \le 2\) is a possible solution region.
- Therefore, we conclude the inequality is satisfied for \(x \in (-\infty, 2]\). This follows from exploring both limits of small and large \(x\) keeping reference to growth comparison.
Thus, the solution set for \(x\) is \((-\infty, 2]\).