Decay rate,k is 2.7% per year=-0.027. What is the half-life in years

Question

i need help understanding word problems in precalculus algebra. 

Answer 1

 

You should know this formula for exponential growth and decay.

 

f(t) = Ae^kt

 

f(t) = value at time "t"

A = the initial/starting value

k = rate of growth (k > 0) or decay(k < 0)

t = time

e = Euler's number, important irrational number used in mathematics, base of natural logarithms

 

Information given:

Decay rate, k is 2.7% per year

k = -0.027.

 

What is the half-life in years?

 

The half-life of a substance undergoing decay is the time (t) it takes for the amount of the substance, (A) to decrease by half.

 

So, the t value when f(t) = A/2

 

A/2 = Ae^-0.027t

 

The initial amount of substance, A cancels out on both sides

 

1/2 = e^-0.027t

 

Use natural logarithms to solve for t

Take the natural log of both sides

 

ln(1/2) = ln(e^-0.027t)

 

Since Euler's number "e" is the base for natural logarithms it cancels out leaving

 

ln(1/2) = -0.027t

 

ln(1/2)/-0.027 = t