We are tasked with solving for $x$ in the equation:

$\log_2(5x) + 4 \log_2(3) = 5$

Let's break down the steps to solve it:

Step 1: Simplify the second term

Using the logarithmic rule $a \log_b(c) = \log_b(c^a)$, we can simplify the second term:

$4 \log_2(3) = \log_2(3^4) = \log_2(81)$

Now the equation becomes:

$\log_2(5x) + \log_2(81) = 5$

Step 2: Combine the logarithmic terms

Using the logarithmic rule $\log_b(a) + \log_b(c) = \log_b(ac)$, we can combine the two logarithmic terms on the left-hand side:

$\log_2(5x \cdot 81) = 5$

This simplifies to:

$\log_2(405x) = 5$

Step 3: Solve for $x$

To eliminate the logarithm, we rewrite the equation in exponential form:

$405x = 2^5$

Since $2^5 = 32$, we have:

$405x = 32$

Solving for $x$:

$x = \frac{32}{405}$

Final Answer:

The exact value of $x$ is:

$x = \frac{32}{405}$

Would you like more details on any specific step? Feel free to ask! Here are some follow-up questions you might find useful:

1. What are the properties of logarithms that help in solving logarithmic equations?
2. How can logarithms with different bases be handled?
3. Why is it useful to rewrite logarithmic equations in exponential form?
4. Can the solution for $x$ in this equation be checked by substitution?
5. What other methods exist for solving logarithmic equations like this?

Tip: When solving logarithmic equations, always look for opportunities to combine logarithmic terms using properties like the product and power rules.