Suppose a quantity is halvd every 18 years. use the approximate half-life formula to estimate its decay rate.
is the half life equation
y= 1/2^t1/2
2^1/2t= 1/y
t 1/2= log2 (1/y)
Where do I put 18 at?
quantity=originalamount*2-t/18
or as you prefer
quantity=originalamount*(1/2)t/18
Where do you get the original amount and the quantity from? I am confused since the problem doesnt have anymore numbers than 18
I really don't know what the question is asking, unless it wants the "decay rate"
quantity=oritinal amount(2t/18
take the ln of each side
ln(quantity)=ln(original)+t/18 * ln 2
now take the deravative...
1/q dq/dt=0+1/18 *ln2
so the decay rate is
dq/dt=q*ln2 / 18
What does dq stand for and dt? sorry for all the questions
dq/dt is rate of q changing with respect to time. It is a calulus term. dx/dt is rate of change of x, and so on. That is what surprised me about the question, rate is a calculus term.
To estimate the decay rate of a quantity that is halved every 18 years using the approximate half-life formula, you need to incorporate the value of 18 into the formula.
The approximate half-life formula is given by:
t1/2 = log2(1/y)
Where:
- t1/2 is the half-life of the quantity (time it takes for the quantity to be halved)
- y is the decay factor (how much the quantity is reduced)
In this case, the quantity is halved every 18 years, so the decay factor (y) is 1/2.
To find the decay rate using the approximate half-life formula, substitute the decay factor (y) into the formula:
t1/2 = log2(1/(1/2))
Simplifying this expression gives:
t1/2 = log2(2)
Since log2(2) equals 1, the equation becomes:
t1/2 = 1
Thus, the decay rate of the quantity that is halved every 18 years can be estimated as 1 unit of time (which could be years or any other time unit).
In summary, to estimate the decay rate of a quantity halved every 18 years using the approximate half-life formula, you substitute the value of the decay factor (y) into the formula, which in this case gives a decay rate of 1 unit of time.