To solve the equation:

$2^{(x + 1)} = 3^{(-x + 1)}$

we take the natural logarithm of both sides:

$\ln\left(2^{(x+1)}\right) = \ln\left(3^{(-x+1)}\right)$

By applying the logarithmic rule $\ln(a^b) = b \ln(a)$, this simplifies to:

$(x+1)\ln(2) = (-x+1)\ln(3)$

Now, expand both sides:

$x\ln(2) + \ln(2) = -x\ln(3) + \ln(3)$

Next, collect the terms involving $x$ on one side:

$x\ln(2) + x\ln(3) = \ln(3) - \ln(2)$

Factor out $x$:

$x(\ln(2) + \ln(3)) = \ln(3) - \ln(2)$

Solve for $x$:

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

Now, we can calculate the value of $x$. Let me do that.The value of $x$ is approximately $0.23$.

Therefore, the correct answer is:

$x \approx 0.23$

Do you want any further details or explanations?

Here are 5 related questions to expand on this topic:

1. How do you solve logarithmic equations with different bases?
2. What is the change of base formula for logarithms?
3. Can you explain how to handle exponential equations where the exponents contain variables?
4. What are some common logarithmic properties used in equation solving?
5. How does the natural logarithm ($\ln$) differ from other logarithms?

Tip: Remember that logarithmic functions are the inverse of exponential functions!