From the equation e^(-0.055x) = 2e^(-0.124x), we can first take the natural logarithm of both sides: ln(e^(-0.055x)) = ln(2e^(-0.124x)) ln(e^(-0.055x)) = ln2+ln(e^(-0.124x)) Since ln(e^a) = a, -0.055x = ln2 - 0.124x 0.069x = ln2 x = 10.05, or approximately 10 days