Problem 24 Solution

dc = 100-c dt

dc 1/(100-c) = dt

-ln(100-c) = t + K₁

t = ln(K₂) - ln(100-c) (where K₁ = ln(K₂)

t = ln(K₂/(100-c))

When t=0 c=15 so K₂ must be 85

t = ln(85/(100-c))

Solving for t when c=25:

t = ln(85/(100-25)) = ln(85/75) =~ ln(17/15) = 0.1252