Question 821773: The effectiveness of a drug is given by the formula 
Where N is the number of effective particles and t is the time in years.
Plot a graph N against t for values of t from 0 to 5 years
Determine by calculation the half life of the drug (the time it takes for the drug to have half the original number of effective particles)
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website!
(Note: Since fractional exponents do not display well on algebra.com, I am going to replace with its decimal equivalent:
I really can't help much with the graph. Here are some suggestions:- If you have a graphing calculator, then enter the function into your calculator to see what the graph look like. (If your calculator does not have a button for "e" or
 , then replace the "e" in the function with 2.71828183 (or some rounded-off version of this number). - With or without a graphing calculator, the problem suggests that you find specific points for t = 0, 1, 2, 3, 4 and 5. (Again, use 2.71828183 (or some rounded-off version of this number) if your calculator does not have a button for "e" or ).
t = 0 represents a starting point. If you put a zero in for the "t" in the function, the exponent becomes zero, too. And "e" to the zero power (like every other non-zero number to the zero power) is 1. So N = 250 when t = 0. This represents the initial number of particles. To find the half-life we will need to find out how long it takes for the number of particles to go down to half of 250 (i.e. 125). To find this amount of time we replace the N with 125 and solve:
To solve this, we start by isolating the base and its exponent. Dividing both sides by 250:
which simplifies to:
Next we use logarithms. Any base of logarithm may be used. But there are advantages to choosing certain bases:- Choosing a base for the logarithm that matches the base of the exponent will result in the simplest possible exact expression for the solution.
- Choosing a base for the logarithm that your calculator "knows", base 10 ("log") or base e ("ln"), will result in an expression for the solution that can easily be converted to a decimal approximation if one is needed.
With our equation we can get both advantages by choosing to use base e logarithms:
Next we use a property of logarithms,  , which allows us to "move" the exponent of the argument out in front of the log. (It is this very property of logs that is the reason we use logs on equations like this. It allows us to move the exponent, where the variable is, out in front where we can "get at" the variable and solve for it.) Using this property on the right side of the equation we get:
By definition, ln(e) = 1. (This is why matching the bases leads to simpler expressions.) So this simplifies to:
And finally we divide by -0.25:
This is an exact expression for the solution. Since it is awkward to say that the half-life is  years, we will use our calculator to get a decimal approximation:
 (rounded to four places)
2.7724 = t
So the half-life is approximately 2.8 years.
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