To solve for \( x \) in the equation

\[
\frac{6^a \cdot 2^a}{12^b} = 12^x,
\]

we can follow these steps:

### Step 1: Simplify the left side
First, express the bases on the left side in terms of prime factors:
- \( 6 = 2 \times 3 \)
- \( 12 = 2^2 \times 3 \)

Thus, the equation becomes:

\[
\frac{(2 \times 3)^a \cdot 2^a}{(2^2 \times 3)^b} = 12^x
\]

### Step 2: Expand and combine the terms
Expand the terms on the left-hand side:
\[
\frac{2^a \cdot 3^a \cdot 2^a}{2^{2b} \cdot 3^b} = \frac{2^{2a} \cdot 3^a}{2^{2b} \cdot 3^b}
\]

### Step 3: Combine like terms
Now, simplify by combining the powers of the same base:
\[
\frac{2^{2a}}{2^{2b}} \cdot \frac{3^a}{3^b} = 2^{2a - 2b} \cdot 3^{a - b}
\]

The equation becomes:
\[
2^{2a - 2b} \cdot 3^{a - b} = 12^x
\]

### Step 4: Express \( 12^x \) in terms of primes
Recall that \( 12^x = (2^2 \times 3)^x = 2^{2x} \times 3^x \).

### Step 5: Equate the exponents
For the equation \( 2^{2a - 2b} \cdot 3^{a - b} = 2^{2x} \times 3^x \) to hold true, the exponents of 2 and 3 must be equal on both sides:

- For the base 2: \( 2a - 2b = 2x \)
- For the base 3: \( a - b = x \)

### Step 6: Solve for \( x \)
From \( a - b = x \), we already have:
\[
x = a - b
\]

Substituting \( x = a - b \) into the equation \( 2a - 2b = 2x \), we get:
\[
2a - 2b = 2(a - b)
\]
which confirms that the equation is consistent.

So, \( x = a - b \).

Would you like a more detailed explanation or have any other questions?

Here are 5 related questions:

1. Solve \(\frac{8^a}{4^b} = 16^x\) for \(x\).
2. What is \(x\) in \(\frac{9^a \cdot 3^a}{27^b} = 3^x\)?
3. Solve \( \frac{10^a \cdot 2^a}{20^b} = 10^x\) for \(x\).
4. How do you simplify \(\frac{4^a \cdot 2^a}{8^b} = 2^x\)?
5. Find \(x\) in \(\frac{18^a \cdot 3^a}{54^b} = 18^x\).

**Tip:** When solving equations involving exponents, always express all terms with the same base before equating the exponents.

Learn how to solve the equation (6^a * 2^a) / 12^b = 12^x step-by-step. This problem involves exponents and prime factorization, with detailed explanations and solutions. Suitable for students in Grades 9-12.

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