You need to solve the equation using the laws of exponentials such that:

`-3+e^(x+1)=2+e^(x-2) =gt e^(x+1) - e^(x-2) = 2+3 `

`e^x*e - e^x*e^(-2) = 5 `

You need to factor out `e^x`  such that:

`e^x(e - 1/e^2) = 5` 

`e^x = 5/(e - 1/e^2)`

You need to take logarithms such that:

`ln e^x = ln(5/(e - 1/e^2))`

`x*ln e =ln(5/(e - 1/e^2))`

You need to remember that ln e = 1 such that:

`x =ln((5e^2)/(e^3 - 1))`

`x = ln 5e^2 - ln(e^3 - 1)`

`x = ln 5 + 2 ln e - ln(e^3 - 1)`

`x = ln 5 + 2 - ln(e^3 - 1)`

Since `e^3 gt 1` , hence `e^3 - 1 gt 0` , thus it is possible to evaluate `ln(e^3 - 1).`

Hence, evaluating the solution to the given equation yields x = `ln 5 + 2 - ln(e^3 - 1).`

`-3+e^(x+1)=2+e^(x-2)` 

`e^(x+1)+e^(x-2)=5`