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.