To solve the exponential equation $4^{(5x + 1)} = 5^{(x - 4)}$, we will use logarithms to express the solution exactly. Here's the step-by-step process:

Step 1: Take the natural logarithm on both sides

Apply the natural logarithm (ln) to both sides of the equation:

$\ln \left( 4^{(5x + 1)} \right) = \ln \left( 5^{(x - 4)} \right)$

Step 2: Use the logarithmic power rule

Using the logarithmic property $\ln(a^b) = b \ln(a)$, rewrite the equation:

$(5x + 1) \ln(4) = (x - 4) \ln(5)$

Step 3: Expand both sides

Now expand the terms on both sides:

$5x \ln(4) + \ln(4) = x \ln(5) - 4 \ln(5)$

Step 4: Group like terms

Move all terms involving $x$ to one side and the constants to the other side:

$5x \ln(4) - x \ln(5) = -4 \ln(5) - \ln(4)$

Step 5: Factor out $x$

Factor $x$ from the left-hand side:

$x \left( 5 \ln(4) - \ln(5) \right) = -4 \ln(5) - \ln(4)$

Step 6: Solve for $x$

Finally, divide both sides by $5 \ln(4) - \ln(5)$ to solve for $x$:

$x = \frac{-4 \ln(5) - \ln(4)}{5 \ln(4) - \ln(5)}$

This is the exact solution in terms of logarithms.

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Would you like any further explanation or additional details?

Here are 5 related questions to expand on this:

1. What properties of logarithms can be useful in simplifying exponential equations?
2. How can this method be adapted to equations with different bases on both sides?
3. Can logarithms be used to solve exponential equations with more than two terms?
4. How do logarithms relate to exponential growth and decay in real-world scenarios?
5. What is the significance of the natural logarithm compared to logarithms of other bases?

Tip: Remember, the natural logarithm (ln) is especially useful in solving equations involving the number $e$, but it can be applied to any base!