We are given the equation:

$\log_2(5x) - \log_4(2x) = 3$

Step 1: Express $\log_4(2x)$ in terms of base 2

We can use the change of base formula:

$\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$

In this case, converting $\log_4(2x)$ to base 2:

$\log_4(2x) = \frac{\log_2(2x)}{\log_2(4)} = \frac{\log_2(2x)}{2}$

Thus, the equation becomes:

$\log_2(5x) - \frac{1}{2} \log_2(2x) = 3$

Step 2: Simplify the logarithms

Next, apply the logarithmic identity $\log_b(ab) = \log_b(a) + \log_b(b)$:

$\log_2(5x) = \log_2(5) + \log_2(x)$ $\log_2(2x) = \log_2(2) + \log_2(x) = 1 + \log_2(x)$

Substituting these into the equation:

$\left(\log_2(5) + \log_2(x)\right) - \frac{1}{2}(1 + \log_2(x)) = 3$

Step 3: Solve for $\log_2(x)$

Distribute the $\frac{1}{2}$ in the second term:

$\log_2(5) + \log_2(x) - \frac{1}{2} - \frac{1}{2}\log_2(x) = 3$

Combine like terms:

$\log_2(5) + \left(1 - \frac{1}{2}\right)\log_2(x) - \frac{1}{2} = 3$

Simplify further:

$\log_2(5) + \frac{1}{2} \log_2(x) - \frac{1}{2} = 3$

Now, isolate $\log_2(x)$:

$\frac{1}{2} \log_2(x) = 3 + \frac{1}{2} - \log_2(5)$

$\frac{1}{2} \log_2(x) = \frac{7}{2} - \log_2(5)$

Multiply both sides by 2:

$\log_2(x) = 7 - 2\log_2(5)$

Step 4: Solve for $x$

Exponentiate both sides to remove the logarithm:

$x = 2^{7 - 2\log_2(5)} = 2^7 \cdot 2^{-2\log_2(5)}$

Since $2^{\log_2(a)} = a$, this becomes:

$x = 128 \cdot 5^{-2}$

$x = 128 \cdot \frac{1}{25} = \frac{128}{25}$

Thus, the solution is:

$x = \frac{128}{25}$

Let me know if you'd like further clarification on any part of this process!

Related Questions:

1. How do you change the base of a logarithm?
2. What is the logarithmic identity for $\log_b(ab)$?
3. How do you simplify logarithmic expressions with coefficients?
4. How can you solve logarithmic equations using properties of exponents?
5. What is the purpose of the change of base formula?

Tip:

When dealing with logarithms of different bases, converting everything to a common base simplifies the process tremendously.