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We are tasked with solving the following limit:

\[
\lim_{x \to \infty} \frac{\ln(10^x + e^{2x})}{x}
\]

### Step-by-Step Solution:

1. **Examine the expression inside the logarithm:**

   \[
   10^x + e^{2x}
   \]

   As \( x \to \infty \), \( e^{2x} \) grows much faster than \( 10^x \). Therefore, for very large \( x \), the term \( e^{2x} \) dominates, and we can approximate the expression inside the logarithm:

   \[
   10^x + e^{2x} \approx e^{2x}
   \]

2. **Logarithmic simplification:**

   Now apply the approximation to the logarithm:

   \[
   \ln(10^x + e^{2x}) \approx \ln(e^{2x}) = 2x
   \]

3. **Simplify the ratio:**

   Now substitute the approximation into the original limit:

   \[
   \lim_{x \to \infty} \frac{\ln(10^x + e^{2x})}{x} \approx \lim_{x \to \infty} \frac{2x}{x} = 2
   \]

Thus, the value of the limit is:

\[
\boxed{2}
\]

---

Would you like further details or have any other questions?

Here are five related questions to explore:

1. How does the approximation \( e^{2x} \) dominate over \( 10^x \) as \( x \to \infty \)?
2. What is the behavior of logarithmic functions for large inputs, and why does \( \ln(e^{2x}) = 2x \)?
3. Can L'Hopital's rule be applied to this limit, and if so, how?
4. What are other common functions that grow faster than \( 10^x \)?
5. How would the limit change if the base of the exponential terms were different?

**Tip:** When dealing with limits at infinity, comparing the growth rates of terms in the expression is often a key strategy for simplifying complex functions.

Find the limit: \lim_{x \to \infty} \frac{\ln(10^x + e^{2x})}{x}.

Explore how to solve the limit \lim_{x \to \infty} \frac{\ln(10^x + e^{2x})}{x} by analyzing the behavior of logarithmic and exponential functions as x approaches infinity. The step-by-step solution uses growth rate comparisons and logarithmic simplifications.

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