3e^(-s-2)e^(2s)=9 How do I solve this problem

in the solutions, it was 3e^(s-2) = 9 but where did the e^(2s) go?
answer is s = ln3 + 2

When you multiply e^(-s-2) and e^2s , you add the exponents and get

e^(-s-2) * e^2s = e^[ -s - 2 + 2s] = e^(s - 2)

so

3e^(-s-2)e^(2s) = 3e^(s-2)

The problem simplifies to

3e^(s-2) = 9

Divide both sides by 3 to get

e^(s-2) = 3

ln [ e^(s-2)] = ln(3) [the application of the natural log on the left gets rid of the natural base 'e']

s - 2 = ln(3)

s = ln(3) + 2