The half life of a substance is the time it takes for 1/2 of the substance to decay, metabolize, or be used up. For example, if 1/2 of a drug is metabolized in 3 hours, after 6 hours 1/4 of the drug is left and 3/4 of the drug has been used:



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Click on the blue points on the sliders and drag them to change the figure. What changes when D_0 changes? What changes when h changes? |
| Manipulative 1 - Half Life Created with GeoGebra. |
The half life of a radioactive substance is 3 hours. The initial amount is 3 grams. How long before only 0.6 grams is left?
| Step | Equation | Description |
|---|---|---|
| 1 | ![]() |
Start with the half-life formula. |
| 2 | ![]() |
Fill the values into the formula. The initial amount D0 = 3. The half-life h = 3. The amount left after t hours is D = 0.6. Solve for t. |
| 3 | ![]() |
Use the multiplication property of equality to multiply both sides of the equation by 1/3. |
| 4 | ![]() |
Use the logarithm to convert the equation from exponential form to logarithmic form. The definition of a logarithm is logab = c if and only if ac = b. In this case a = 1/2, b = 0.2 and c = t/3. |
| 5 | ![]() |
Use the multiplication property of equality to multiply both sides of the equation by 3. |
| 6 | ![]() |
Use the
change of base formula
to convert the logarithm to base 10. The change of base formula is
. In this case,
a = 1/2,
x = 0.2 and
b = 10. |
| 7 | ![]() |
Substitute the values of the logarithms into the equation. log100.2 ≈ -0.69897. log100.5 ≈ -0.30103. |
| 8 | t ≈ 6.9658 | Calculate the approximate value of t. After about 7 hours there will only be 0.6 grams left. |
| Table 2: Half life example. | ||
| # | A | B | C | D |
| E | F | G | H | I |
| J | K | L | M | N |
| O | P | Q | R | S |
| T | U | V | W | X |
| Y | Z |
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