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Let's analyze the problem step by step.

The given system of equations is:

\[
2^n = 5 \quad \text{and} \quad 3^n = 4
\]

We are asked to find the value of:

\[
6^{n-1}
\]

### Step 1: Solve for \(n\)
We can use logarithms to find \(n\) by solving the first equation:

\[
2^n = 5
\]

Taking the natural logarithm on both sides:

\[
n \ln(2) = \ln(5)
\]

Thus, \(n\) is:

\[
n = \frac{\ln(5)}{\ln(2)}
\]

Similarly, from the second equation \(3^n = 4\), we get:

\[
n \ln(3) = \ln(4)
\]

Thus,

\[
n = \frac{\ln(4)}{\ln(3)}
\]

### Step 2: Equate the two expressions for \(n\)
We now have two expressions for \(n\):

\[
\frac{\ln(5)}{\ln(2)} = \frac{\ln(4)}{\ln(3)}
\]

This equation allows us to solve for the value of \(n\) in the context of logarithmic relationships between these numbers. However, instead of focusing on finding the exact value of \(n\), we can use these relationships to help simplify the expression for \(6^{n-1}\).

### Step 3: Simplify \(6^{n-1}\)
We know that:

\[
6^{n-1} = (2 \cdot 3)^{n-1} = 2^{n-1} \cdot 3^{n-1}
\]

Using the given information \(2^n = 5\) and \(3^n = 4\), we can express \(2^{n-1}\) and \(3^{n-1}\) as:

\[
2^{n-1} = \frac{2^n}{2} = \frac{5}{2}
\]

\[
3^{n-1} = \frac{3^n}{3} = \frac{4}{3}
\]

### Step 4: Multiply the results
Now, multiply these two expressions:

\[
6^{n-1} = \frac{5}{2} \cdot \frac{4}{3} = \frac{20}{6} = \frac{10}{3}
\]

Thus, the value of \(6^{n-1}\) is:

\[
\boxed{\frac{10}{3}}
\]

Would you like more details on this, or have any questions? Here are some related questions to explore:

1. What properties of logarithms can be applied to simplify exponential equations?
2. How do logarithms help solve equations involving exponential terms?
3. How could we solve similar equations if the bases were not easily related?
4. How would the solution change if \(6^{n-1}\) had a different base, such as 5 instead of 6?
5. Can we use this method to solve exponential equations with irrational or transcendental numbers?

**Tip:** When solving exponential equations, always consider taking logarithms on both sides to bring down the exponents and simplify the equations.

Solve for 6^(n-1) given that 2^n = 5 and 3^n = 4.

Learn how to solve the equation 6^(n-1) given that 2^n = 5 and 3^n = 4. This problem covers key algebraic concepts such as logarithms and exponent laws, and includes a step-by-step solution suitable for high school students.

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