To solve the equation:

\[
e^{x+1} = 5^{x-2}
\]

We can follow these steps:

### Step 1: Apply the natural logarithm (ln) to both sides.
Taking the natural logarithm on both sides helps to deal with the exponents:

\[
\ln(e^{x+1}) = \ln(5^{x-2})
\]

### Step 2: Use the properties of logarithms.
We use the logarithmic identity \(\ln(a^b) = b \ln(a)\) to simplify both sides:

\[
(x+1) \ln(e) = (x-2) \ln(5)
\]

Since \(\ln(e) = 1\), the equation simplifies to:

\[
x+1 = (x-2)\ln(5)
\]

### Step 3: Expand and simplify the equation.
Distribute \(\ln(5)\) on the right-hand side:

\[
x+1 = x \ln(5) - 2 \ln(5)
\]

### Step 4: Isolate \(x\).
Move all terms involving \(x\) to one side:

\[
x - x \ln(5) = -1 - 2 \ln(5)
\]

Factor out \(x\) on the left-hand side:

\[
x(1 - \ln(5)) = -1 - 2 \ln(5)
\]

### Step 5: Solve for \(x\).
Finally, solve for \(x\) by dividing both sides:

\[
x = \frac{-1 - 2 \ln(5)}{1 - \ln(5)}
\]

This is the exact solution for \(x\).

If you'd like, I can also compute the approximate numerical value of \(x\).

Would you like more details or the numeric approximation?

This guide provides a detailed step-by-step solution to solving the exponential equation e^(x+1) = 5^(x-2). It covers key mathematical concepts such as logarithms, properties of exponents, and algebraic manipulation. Suitable for high school students in Grades 10-12.

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