Use applications of exponential functions and logarithmic functions to solve.
The half-life of plutonium 241 is 14.4 years.If 100 grams is present now, How much will it take to reach 2 grams of plutonium 241?
y=a(1/2)^(t/14.4)
How do I solve for t? I'm not sure where to even begin.
100(1/2)^(t/14.4) = 2
(1/2)^(t/14.4) = 0.02
t/14.4 ln(1/2) = ln(0.02)
t/14.4 = ln(0.02)/ln(0.5)
That is just log0.50.02
t = 14.4 ln(0.02)/ln(0.5) = 81.27
That's about 5.6 half-lives. That makes sense, since
(1/2)^5 = 1/32 and
(1/2)^6 = 1/64
Your fraction (1/50) is between those two values.
To solve for t in the equation y = a(1/2)^(t/14.4), where y represents the final amount (2 grams), a represents the initial amount (100 grams), and 14.4 represents the half-life of plutonium 241, you can use logarithmic functions.
1. Start by rearranging the equation to isolate the exponential term:
(1/2)^(t/14.4) = y/a
2. Take the logarithm (base 1/2) of both sides to eliminate the exponent on the left side:
log[(1/2)^(t/14.4)] = log(y/a)
3. Apply the logarithm property log(base a) (a^n) = n, which allows you to bring the exponent down as a coefficient:
(t/14.4)*log(1/2) = log(y/a)
4. Rewrite log(1/2) as a negative exponent and simplify:
(t/14.4)*(-log2) = log(y/a)
5. Divide both sides of the equation by -log2 to solve for t:
t/14.4 = log(y/a)/(-log2)
6. Multiply both sides of the equation by 14.4 to isolate t:
t = 14.4 * log(y/a)/(-log2)
By plugging in the given values for y (2 grams) and a (100 grams), you can solve for t using a calculator or logarithmic tables.