SOLUTION: PROBLEM: The amount of a radioactive tracer remaining after 't' days is given by {{{A=A0e^(-0.058t)}}}, where A0 is the starting amount at the beginning of the time period. How

Algebra ->  Trigonometry-basics -> SOLUTION: PROBLEM: The amount of a radioactive tracer remaining after 't' days is given by {{{A=A0e^(-0.058t)}}}, where A0 is the starting amount at the beginning of the time period. How       Log On


   

Question 80072This question is from textbook Algebra and Trigonometry with Analytic Geometry
: PROBLEM:
The amount of a radioactive tracer remaining after 't' days is given by A=A0e%5E%28-0.058t%29, where A0 is the starting amount at the beginning of the time period. How many days will it take for one half of the original amount to decay?

OPTIONS:
a. 10 days
b. 11 days
c. 12 days
d. 13 days

MY WORK SO FAR:
FORMULA: "Law of growth or decay" let 'A0' be the value of a quantity 'A' at time 't'=0 (that is, 'A0' is the initial amount of 'A'.) If 'A' changes instantaneously at a rate proportional to itscurrent value, then A=A%28t%29=A0e%5Ert where r>0 is the rate of growth (or r<0 is the rate of decay) of A.
therefore if we use the formula then we have
A=A0%282.71828%29%5E%28-0.058t%29
QUESTIONS:
1. What is the first step into solving this problem?
2. What are the ways to gather the information needed to continue solving this problem? This question is from textbook Algebra and Trigonometry with Analytic Geometry

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
A=A%5B0%5De%5E%28-0.058t%29
.
Let's just assume that the original amount A%5B0%5D is 1. Then to decay by half the amount
the resulting value of A would be 1%2F2. Substitute these two values and you
get:
.
1%2F2=1%2Ae%5E%28-0.058t%29
.
and after the multiplication by 1, the right side becomes just:
.
1%2F2=e%5E%28-0.058t%29
.
Because an exponent contains the variable we are to solve for, that's a clue that we can use
logarithms and the rules of logarithms to solve. And because e is also involved, let's
take the natural log of both sides to get:
.
ln%281%2F2%29+=+ln%28e%5E%28-0.058t%29%29
.
A rule of logarithms that can be applied is that if the quantity that the logarithm
acts on has an exponent, you can make that exponent a multiplier of the logarithm.
Applying this rule to our problem results in:
.
ln%281%2F2%29+=+-0.058t%2Aln%28e%29
.
But the natural logarithm of e is equal to 1. So substitute 1 for ln%28e%29 to get:
.
ln%281%2F2%29+=+-0.058t+%2A+1
.
Simplify the right side by doing the multiplication and get:
.
ln%281%2F2%29+=+-0.058t
.
Now you can use a calculator to find the natural logarithm of 1%2F2 or its equivalent 0.5.
This logarithm is -0.69314718 and when this is substituted for ln%281%2F2%29 the equation
becomes:
.
+-0.69314718+=+-0.058t
.
Finally solve for t by dividing both sides by -0.058 to get:
.
t+=+%28-0.69314718%29%2F%28-0.058%29+=+11.95081346
.
Therefore, the closest answer in your list of possible answers is 12 days. Hope this helps
you to see a way to solve problems such as these.