The problem asks us to demonstrate that:

$\log_5\left(4^x + 3^x\right) < x, \quad \text{for any real } x > 2.$

Step-by-step solution:

1. Analyze the inequality: $\log_5\left(4^x + 3^x\right) < x$ Recall the logarithmic function $\log_b(y)$ is increasing for $b > 1$. Here, the base is $b = 5 > 1$, so the inequality holds if: $4^x + 3^x < 5^x$

2. Rewrite the inequality: Divide the entire inequality $4^x + 3^x < 5^x$ by $3^x$ (valid for $x > 2$ since $3^x > 0$): $\left(\frac{4}{3}\right)^x + 1 < \left(\frac{5}{3}\right)^x$

3. Define key terms: Let $a = \frac{4}{3}$ and $b = \frac{5}{3}$, where $1 < a < b$. The inequality becomes: $a^x + 1 < b^x$

4. Behavior of exponential functions for $x > 2$:

   • Since $a < b$, $b^x$ grows faster than $a^x$ as $x$ increases.
   • For $x > 2$, the term $a^x + 1$ is strictly smaller than $b^x$.

5. Verification for $x = 2$: When $x = 2$, $4^2 + 3^2 = 16 + 9 = 25, \quad \text{and } 5^2 = 25.$ Thus, equality holds at $x = 2$. For $x > 2$, the inequality becomes strict because $b^x$ grows faster than $a^x + 1$.

6. Conclusion: Therefore, for $x > 2$, $\log_5\left(4^x + 3^x\right) < x$

Let me know if you'd like more details or explanations!

Related Questions:

1. What is the general behavior of exponential functions for different bases?
2. How do logarithmic functions behave with increasing arguments?
3. Why is dividing by $3^x$ valid in this case?
4. Can this inequality hold for $x \leq 2$?
5. How does the base of the logarithm (here 5) influence the inequality?

Tip:

When working with logarithmic inequalities, always ensure the argument of the logarithm is positive and check the monotonicity of the logarithmic function!