To solve the equation $3^{x+1} = 5^{2x-1}$ for $x$, let's proceed by taking the logarithm of both sides:

Step-by-Step Solution:

1. Start with the given equation: $3^{x+1} = 5^{2x-1}$

2. Take the natural logarithm of both sides: $\ln(3^{x+1}) = \ln(5^{2x-1})$

3. Use the power rule of logarithms, $\ln(a^b) = b \ln(a)$: $(x+1) \ln(3) = (2x - 1) \ln(5)$

4. Expand both sides: $x \ln(3) + \ln(3) = 2x \ln(5) - \ln(5)$

5. Group terms involving $x$ on one side and constant terms on the other side: $x \ln(3) - 2x \ln(5) = - \ln(5) - \ln(3)$

6. Factor out $x$ on the left side: $x (\ln(3) - 2 \ln(5)) = - (\ln(5) + \ln(3))$

7. Solve for $x$: $x = \frac{- (\ln(5) + \ln(3))}{\ln(3) - 2 \ln(5)}$

8. Calculate the values of the logarithms:

   • $\ln(3) \approx 1.0986$
   • $\ln(5) \approx 1.6094$

   Substitute these values into the equation:

   $x = \frac{- (1.6094 + 1.0986)}{1.0986 - 2 \cdot 1.6094}$

9. Simplify the calculations: $x = \frac{- 2.708}{1.0986 - 3.2188} = \frac{- 2.708}{-2.1202} \approx 1.28$

Final Answer:

$x \approx 1.28$

Would you like more details on any part of this solution, or do you have further questions?

Here are some related questions that could expand your understanding:

1. How would the solution change if the equation was $3^{x+1} = 5^{x-1}$?
2. Can you solve $3^{x+2} = 4^{2x-3}$ using similar steps?
3. What would happen if we used a different logarithmic base, such as base 10, instead of the natural logarithm?
4. How can we verify the solution using substitution or graphing?
5. What are some common mistakes to watch out for when solving exponential equations?

Tip: When solving equations involving exponents, logarithmic properties like the power rule ($\ln(a^b) = b \ln(a)$) are extremely useful for simplifying the expression.