How do I calculate a time given a starting value, end value, and percentage of change?

Decay at a constant rate follows the formula

A(t) = (A0)e^(kt)

A0 is the initial amount

k is a constant (which will be negative for decay, positive for growth) that describes the rate.

t is the time in some convenient units, such as years.

If it decreases 5% per year, k will just be ln(.95). Then the equation becomes

A(t) = A0(.95^t)

If you have the half-life of some radioactive substance, say 14 years, like Plutonium-241, then the constant is ln(0.5)/14, giving you A0 e^(ln0.5 / 14)t = A0 (0.5)^(t/14). The last expression makes it clear that every 14 years, the amount is cut in half.

If the amount reduces to p percent every y years, then the constant is ln(p)/y.

Note that p is what you have left. It reduces by 1-p every y years. What you need in the constant is the percentage left. When you have the half-life, that's not clear.

Now you can work out your question.

Solve 25 = 35 (0.95^t)

25/35 = 5/7 = 0.95^t = e^ln(.95)t

Take log or ln of each side--it doesn't matter which one you use.

log(5/7) = log5 - log7 = tln(0.95)

t = (log5 - log7) / ln(0.95) = 6.56

Check: 35 x .95^6.56 = 24.99970

BTW, I think the units are strange in your equation. You have ug/m^3, which is a density. Generally decay like this would refer just to grams. Of course, if you had a sample which had a density of 35ug/m^3 of a decaying substance, if the volume doesn't change, you'd end up with a density of 25ug/m^3. So it doesn't really matter, but I thought I would point that out.

Hope this clarifies it. It's a good question, and the formula isn't that complicated if you have a working knowledge of e^x and logarithms.