Growth vs Decay Math Example 2

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Example 2

medium
A radioactive substance has a half-life of 1010 years. Starting with 200200 g, write the decay function and find the amount remaining after 3535 years.

Solution

  1. 1
    Half-life model: A(t)=200⋅(12)t/10A(t) = 200 \cdot \left(\frac{1}{2}\right)^{t/10}.
  2. 2
    At t=35t=35: A(35)=200⋅(0.5)3.5=200⋅123.5=200⋅182=20011.314≈17.68A(35) = 200 \cdot (0.5)^{3.5} = 200 \cdot \frac{1}{2^{3.5}} = 200 \cdot \frac{1}{8\sqrt{2}} = \frac{200}{11.314} \approx 17.68 g.
  3. 3
    Interpretation: in 3535 years (3.53.5 half-lives), the amount decreases from 200200 g to approximately 17.717.7 g.

Answer

A(t)=200⋅(0.5)t/10A(t) = 200\cdot(0.5)^{t/10}; A(35)≈17.7A(35) \approx 17.7 g
Radioactive decay is modeled by A(t)=A0⋅(1/2)t/TA(t)=A_0\cdot(1/2)^{t/T} where TT is the half-life. The base 1/21/2 reflects that each half-life period reduces the quantity by 50%50\%.

About Growth vs Decay

Exponential growth occurs when a quantity multiplies by a factor >1> 1 repeatedly; exponential decay when it multiplies by a factor between 0 and 1.

Learn more about Growth vs Decay →

More Growth vs Decay Examples

Example 1 easy

Classify each function as growth or decay, and find its value at [formula]: (a) [formula], (b) [form

Example 3 easy

A population grows by [formula] per year. Starting at [formula], write the growth function and find

Example 4 hard

Show algebraically that [formula] is an exponential decay function by rewriting it in the form [form

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