Exponents and Logarithms

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  • Topic: Radioactive decay, Exponential function, Exponential decay
  • Pages : 6 (2089 words )
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  • Published : January 6, 2013
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Maryann Crisci
Mrs. Cappiello
Algebra 2/Trig, Period 6
1 April 2012
Exponents and Logarithms
An exponent is the number representing the power a given number is raised to. Exponential functions are used to either express growth or decay. When a function is raised to a positive exponent, it will cause growth. However, when a function is raised to a negative exponent, it will cause decay. Logarithms work differently than exponents. Logarithms represent what power a base should be raised to in order to produce a specific given number. Logarithmic and exponential functions are often used together because they are inverse functions, and therefore “undo” one another. The natural exponential function and the natural logarithm are frequent examples. These functions have a base of the constant e, or Euler’s number. The natural exponential function is written as f(x)=ex. Its inverse, the natural logarithmic function, is written as f(x)=ln(x).

The decay of radioactive substances can be represented by using an exponential function. The mass of the radioactive substance will decrease as time passes, but the rate of decay and the mass of the substance will always remain directly proportional. The decay of radioactive substances can be expressed by the function m(t)=m0e-rt, where m(t) represents the mass remaining at time t, r represents the rate of decay, and m0 represents the initial mass. When the rate of decay is expressed by half-life, the rate of decay expressed as a proportion of the mass can be found by substituting the half-life into the equation ½=1∙e-rh. This is because when h, or the half-life, is equal to t, or time, then the mass of 1 unit becomes ½ unit.

An example of a radioactive substance that will decay over time is Polonium-210. Polonium-210 has a half life of 140 days. This half-life was plugged into the equation ½=1∙e-rh in order to find the rate of decay expressed as a proportion of the mass, or r. 140 was plugged in for h. In order to solve the exponential equation, it is necessary to use the natural logarithm to undo the natural exponential function in the equation. The natural logarithm of each side is taken. The exponents can then be taken out of the equation because the logarithm power law. In other words, the equation can be changed from ln(½)= ln(e)-140r to ln(½)=-140r ln(e). The ln(e) cancels out and then it is necessary to divide both sides of the equation by -140. After doing this, it is found that r≈ .005. Therefore, if the initial mass of a sample of Polonium-210 is 300, then a function that models the amount of Polonium-210 remaining at time t is m(t)=300e-.005t. In order to find the mass remaining after one year, 365 days was plugged into the equation for t. The mass remaining after 1 year, m(365), was approximately 48.37 mg. In order to find the length of time necessary for the sample to decay to 200 mg, 200 was plugged into the same equation for m(t). The natural logarithm was used to cancel out the natural exponent once again. After the equation was solved for t, it was found that it would take approximately 81.09 days for the sample to decay to a mass of 200 mg. The following graph is a graph of the function representing the decay of the 300 mg sample of Polonium-210.

As shown on this graph, as time increases, the mass of the Polonium-210 decreases. Also, as time increases, the rate that the mass decreases will also decrease. This happens because Polonium-210 decays in half-lives. Each time 140 days pass, the substance will have exactly half of the mass it had at the last 140 day interval. When associating this graph with the decay of Polonium-210, only the piece of the graph in quadrant I can be considered. Therefore, the domain is restricted to [0,+∞). This is because it is impossible to have a negative amount of time that has passed. The decay of the substance will start at time 0, and continue as time increases. This also means that the range will be...
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