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.