College Algebra & Trigonometry, 2018a

180 CHAPTER 3. EXPONENTS AND LOGARITHMS
Then take the natural logarithm of both sides and bring the exponent down in
front of the expression as a coefficient:
0.5 =e −kt h
ln(0.5) = ln(e −kt h
)
ln(0.5) = −kt h ∗ ln(e)
ln(0.5) = −kt h ∗ 1
− ln(0.5)
t h
= k
The value of k can then be used in the equation A(t) =A 0 e −kt to determine the
amount of material left after any time t.
Example
The isotope Gold-198 ( 198 Au) is a type of gold sometimes used in medical applications
and has a half-life of 2.7 days. How much of a 65 gram sample of 198 Au
will be left after 6 days? How long would it take for there to be 10 grams left?
If we know the half-life, we can calculate the value of the constant k.
k = − ln(0.5)
t h
k = − ln(0.5)
2.7
k ≈ 0.2567