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To solve the equation 

\[
\log_{2x} (48 \sqrt[3]{3}) = \log_{3x} (162 \sqrt[3]{2})
\]

we will use the property that if \(\log_a (b) = \log_c (d)\), then \(a = c\) and \(b = d\) provided both logarithms are valid.

### Step 1: Simplify the Arguments

First, let's simplify \(48 \sqrt[3]{3}\) and \(162 \sqrt[3]{2}\):

\[
48 \sqrt[3]{3} = 48 \cdot 3^{1/3}
\]

\[
162 \sqrt[3]{2} = 162 \cdot 2^{1/3}
\]

### Step 2: Express Logs in Terms of Common Base

To solve the equation, convert both logarithms to a common base. Let's choose base 10:

\[
\log_{2x} (48 \cdot 3^{1/3}) = \frac{\log_{10} (48 \cdot 3^{1/3})}{\log_{10} (2x)}
\]

\[
\log_{3x} (162 \cdot 2^{1/3}) = \frac{\log_{10} (162 \cdot 2^{1/3})}{\log_{10} (3x)}
\]

Since these are equal,

\[
\frac{\log_{10} (48 \cdot 3^{1/3})}{\log_{10} (2x)} = \frac{\log_{10} (162 \cdot 2^{1/3})}{\log_{10} (3x)}
\]

### Step 3: Simplify the Equations

Simplify the arguments of the logarithms:

\[
\log_{10} (48 \cdot 3^{1/3}) = \log_{10} (48) + \log_{10} (3^{1/3}) = \log_{10} (48) + \frac{1}{3} \log_{10} (3)
\]

\[
\log_{10} (162 \cdot 2^{1/3}) = \log_{10} (162) + \log_{10} (2^{1/3}) = \log_{10} (162) + \frac{1}{3} \log_{10} (2)
\]

So the equation becomes:

\[
\frac{\log_{10} (48) + \frac{1}{3} \log_{10} (3)}{\log_{10} (2x)} = \frac{\log_{10} (162) + \frac{1}{3} \log_{10} (2)}{\log_{10} (3x)}
\]

### Step 4: Solve for \(x\)

Equate the denominators and numerators separately to solve for \(x\):

\[
\log_{10} (2x) = \log_{10} (3x)
\]

Since \(\log_{10} (2x) = \log_{10} (3x)\),

\[
2x = 3x
\]

Solving for \(x\),

\[
0 = x(3-2) \implies x = 0
\]

However, since \(x = 0\) does not make sense in the logarithmic domain (logarithms are not defined for zero or negative arguments), we must discard this solution.

Therefore, there are no real values of \(x\) that satisfy the equation.

Would you like more details on any part of the solution or have any other questions? Here are some related questions you might consider:

1. How do we handle logarithmic equations with different bases?
2. What is the change of base formula for logarithms?
3. How do we simplify logarithmic expressions involving roots and powers?
4. What are the restrictions on the arguments of logarithmic functions?
5. How do we solve logarithmic equations with no real solutions?

**Tip:** When dealing with logarithmic equations, always check the domain to ensure the arguments of the logarithms are valid (positive and non-zero).

Learn how to solve the logarithmic equation log base 2x of 48 times the cube root of 3 equals log base 3x of 162 times the cube root of 2 with a detailed step-by-step solution. This problem covers logarithmic equations with different bases, using the change of base formula for logarithms. Suitable for advanced high school students and anyone interested in logarithmic equations.

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