To solve the equation:

\[
6 \cdot e^{2x} = 4^x
\]

### Step 1: Rewrite \(4^x\) in terms of base \(e\)
We know that \(4 = e^{\ln(4)}\), so \(4^x = e^{x \ln(4)}\). Thus, the equation becomes:

\[
6 \cdot e^{2x} = e^{x \ln(4)}
\]

### Step 2: Divide both sides by 6
\[
e^{2x} = \frac{1}{6} \cdot e^{x \ln(4)}
\]

### Step 3: Take the natural logarithm (ln) of both sides
Apply \(\ln\) to both sides to remove the exponentials:

\[
\ln(e^{2x}) = \ln\left(\frac{1}{6} \cdot e^{x \ln(4)}\right)
\]

This simplifies to:

\[
2x = \ln\left(\frac{1}{6}\right) + x \ln(4)
\]

### Step 4: Solve for \(x\)
Rearrange the equation to isolate \(x\):

\[
2x - x \ln(4) = \ln\left(\frac{1}{6}\right)
\]

Factor out \(x\):

\[
x(2 - \ln(4)) = \ln\left(\frac{1}{6}\right)
\]

Finally, solve for \(x\):

\[
x = \frac{\ln\left(\frac{1}{6}\right)}{2 - \ln(4)}
\]

### Step 5: Simplify the result
We can simplify the logarithmic terms:

\[
x = \frac{\ln\left(\frac{1}{6}\right)}{2 - \ln(4)}
\]

That's the exact solution. You can further evaluate it numerically if needed.

Would you like to proceed with a numerical approximation or need further clarification?

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### Related questions:
1. How do we calculate the natural logarithm of a fraction?
2. What does it mean to express numbers in terms of base \(e\)?
3. Can we solve similar equations with different bases (e.g., \(3^x\))?
4. What are some methods to approximate natural logarithms without a calculator?
5. How does logarithmic differentiation apply in calculus?

### Tip:
Always try to rewrite exponential expressions with different bases in terms of the same base (like \(e\)) for easier manipulation.

Step-by-step solution for solving the exponential equation 6 · e^2x = 4^x. Learn how to rewrite terms in base e, apply logarithmic properties, and solve for x. Suitable for high school students in Grades 10-12.

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