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:
After 9 hours, 1/8 of the drug is left: The half life formula is an exponential function: where D is the remaining substance after time t, D_{0} is the initial amount of the substance, h is the half life of the substance, and t is the elapsed time.
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 D_{0} = 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 log_{a}b = c if and only if a^{c} = 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. log_{10}0.2 ≈ -0.69897. log_{10}0.5 ≈ -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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