SOLUTION: b) The half-life of a radioactive substance is 300 years. If the initial amount is q milligrams, then the quantity q(t) remaining after t years is given by q(t) = q2^(kt). Find k

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: b) The half-life of a radioactive substance is 300 years. If the initial amount is q milligrams, then the quantity q(t) remaining after t years is given by q(t) = q2^(kt). Find k      Log On


   

Question 1198954: b) The half-life of a radioactive substance is 300 years. If the initial amount is q milligrams, then the quantity q(t) remaining after t years is given by q(t) = q2^(kt). Find k
Found 3 solutions by josgarithmetic, math_tutor2020, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Strange way of putting it.

q%28t%29=q2%5E%28kt%29

If half life 300 years then
1%2F2=1%2A2%5E%28300k%29
1%2F2=2%5E%28300k%29
Which base do you want? ...
log%28%281%2F2%29%29=300k%2Alog%28%282%29%29
300k=log%28%281%2F2%29%29%2Flog%28%282%29%29
300k=-log%28%282%29%29%2Flog%28%282%29%29
300k=-1
highlight%28k=-1%2F300%29

The function in that form is highlight%28q%28t%29=q2%5E%28%28-1%2F300%29t%29%29.

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

I'll use the variable 'a' as the initial amount, rather than q, since q is already taken in q(t).

Therefore, I'll rewrite this given equation
q(t) = q*2^(kt)
as
q(t) = a*2^(kt)

initial amount = a
half of that is q/2 = 0.5a
which occurs t = 300 years later

This means q(t) = 0.5a when t = 300
In other words, q(300) = 0.5a

q(t) = a*2^(kt)
q(300) = a*2^(k*300)
0.5a = a*2^(300k)
0.5 = 2^(300k)

I divided both sides by 'a' in the jump from step 3 to step 4.
Ultimately the starting amount does not matter.

From here we use logarithms to solve for k.
If the variable is in the trees (aka exponent), then log it down.

The reason for using logs is because of this property
log(A^B) = B*log(A)
which helps us pull down the exponent to isolate the variable.

So,
0.5 = 2^(300k)
Log(0.5) = Log( 2^(300k) )
Log(0.5) = 300k*Log( 2 )
k = Log(0.5)/(300*Log(2))
k = Log(2^(-1))/(300*Log(2))
k = -1*Log(2)/(300*Log(2))
k = -1/300

---------------------------------------

Here's another way to solve for k without using logs.

0.5 = 2^(300k)
1/2 = 2^(300k)
2^(-1) = 2^(300k)

Both sides have the same base.
In order for both sides to be equal, the exponents must be equal.

-1 = 300k
300k = -1
k = -1/300

This type of trick is handy if you aren't too familiar with logs yet.
However, this trick won't always work for any general exponential equation.

For example, the trick won't work if we had say
0.7 = 2^(300k)
which is why I prefer the logarithm approach.

---------------------------------------

Answer: k = -1/300


-1/300 = -0.003333 where the '3's go on forever.
I think it's best to stick to the fraction form since it's most exact.
If your teacher says otherwise, then be sure to follow those instructions.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
b) The half-life of a radioactive substance is 300 years. If the initial amount is q milligrams, then the quantity q(t) remaining after t years is given by q(t) = q2^(kt). Find k
   
    matrix%281%2C3%2C+1%2F2%2C+%22=%22%2C+2%5E%28300k%29%29 ----- Substituting matrix%281%2C3%2C+1%2F2%2C+for%2C+q%28t%29%2Fq%29 and 300 for t
matrix%281%2C3%2C+300k%2C+%22=%22%2C+log+%282%2C+%281%2F2%29%29%29 ----- Converting to LOGARITHMIC form
 300k = - 1 ----- Substituting - 1 for log+%282%2C+%281%2F2%29%29
  highlight_green%28matrix%281%2C3%2C+k%2C+%22=%22%2C+highlight%28%28-+1%29%2F300%29%29%29