We are tasked with solving the equation:

\[
4^x \cdot 3^{x+1} = 8
\]

### Step 1: Express \(4^x\) and 8 as powers of 2
We know that:
- \(4 = 2^2\), so \(4^x = (2^2)^x = 2^{2x}\).
- \(8 = 2^3\).

Thus, the equation becomes:

\[
2^{2x} \cdot 3^{x+1} = 2^3
\]

### Step 2: Divide both sides by \(2^{2x}\)
Next, isolate the terms involving \(3\). Divide both sides by \(2^{2x}\):

\[
3^{x+1} = \frac{2^3}{2^{2x}} = 2^{3 - 2x}
\]

### Step 3: Convert to logarithmic form
We now have the equation:

\[
3^{x+1} = 2^{3-2x}
\]

Taking the natural logarithm (ln) of both sides:

\[
\ln(3^{x+1}) = \ln(2^{3-2x})
\]

Using logarithmic properties:

\[
(x+1) \ln(3) = (3 - 2x) \ln(2)
\]

### Step 4: Solve for \(x\)
Now expand and solve for \(x\):

\[
x \ln(3) + \ln(3) = 3 \ln(2) - 2x \ln(2)
\]

Group like terms:

\[
x \ln(3) + 2x \ln(2) = 3 \ln(2) - \ln(3)
\]

Factor out \(x\):

\[
x (\ln(3) + 2 \ln(2)) = 3 \ln(2) - \ln(3)
\]

Finally, solve for \(x\):

\[
x = \frac{3 \ln(2) - \ln(3)}{\ln(3) + 2 \ln(2)}
\]

This is the exact solution. For a numerical approximation:

\[
x \approx 0.129
\]

Would you like further details or explanations?

### 5 Related Questions:
1. How are logarithmic properties used in solving exponential equations?
2. Can we use any other base for logarithms when solving such equations?
3. How does dividing by a power of 2 simplify exponential equations?
4. What is the role of factoring in solving linear equations involving logarithms?
5. How can you check if the calculated value of \(x\) satisfies the original equation?

**Tip**: When solving exponential equations, expressing terms in a common base can simplify the process greatly.

Learn how to solve the exponential equation (4^x)3^(x+1) = 8 using logarithmic properties and powers of 2. This detailed solution involves steps like converting terms into powers of 2 and using natural logarithms, suitable for high school students in Grades 10-12.

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