Question

If \(3^x + 2^{2x} \ge 5^x\), then the solution set for \(x\) is

• A. \((-\infty, 2]\)
• B. \([2, \infty)\)
• C. \([0, 2]\)
• D. \([2)\)

Hint:

\((3/5)^x\) and \((4/5)^x\) are decreasing functions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Divide by \(5^x\): \(\left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x \ge 1\). 
Step 2: Detailed Explanation: 
Let \(a = (3/5)^x\), \(b = (4/5)^x\) 
\(a + b \ge 1\) 
Since \((3/5)^x\) and \((4/5)^x\) are decreasing functions, 
check \(x = 2\): \((9/25) + (16/25) = 1\), equality 
For \(x<2\): both terms larger, sum \(> 1\) 
For \(x>2\): sum \(< 1\) 
So \(x \le 2\) 
Step 3: Final Answer: 
\((-\infty, 2]\).