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Zorluk: Çok zorExponential Functions and Equations

A sample of a radioactive isotope decays such that its mass, in grams, is modeled by the function M(t)=Adt12M(t) = A \cdot d^{\frac{t}{12}}, where tt is the time in hours since the measurement began, and AA and dd are positive constants. The mass of the sample decreases by 75%75\% every 2424 hours. If the mass of the sample is 1515 grams when t=36t = 36, what was the initial mass, in grams, of the sample?

Cevap: 120 grams

Cevap

The initial mass of the sample was 120 grams.
The correct answer is 120. A decrease of 75%75\% over 24 hours means the mass at t+24t+24 is 0.250.25 times the mass at tt. According to the function, M(t+24)=Ad(t+24)/12=Adt/12d2=M(t)d2M(t+24) = A \cdot d^{(t+24)/12} = A \cdot d^{t/12} \cdot d^2 = M(t) \cdot d^2. Equating the two yields d2=0.25d^2 = 0.25, so d=0.5d = 0.5. Substituting t=36t = 36 and M(36)=15M(36) = 15 gives 15=A(0.5)36/12=A(0.5)3=0.125A15 = A \cdot (0.5)^{36/12} = A \cdot (0.5)^3 = 0.125A. Solving for AA gives A=120A = 120.

Adım Adım Çözüm

1
Relate the 24-hour decay rate to the exponent in the function to set up an equation for dd.
d2=0.25d^2 = 0.25
Every 24 hours (tt increases by 24), the mass decreases by 75%75\%, so it becomes 25%25\% (0.250.25) of its previous value. The exponent increases by 24/12=224/12 = 2, multiplying the mass by d2d^2.
2
Solve for the decay base dd.
d=0.5d = 0.5
Since dd is a positive constant, taking the square root of 0.250.25 gives 0.50.5.
3
Substitute the given mass at t=36t = 36 into the model to solve for the initial mass AA.
A=120A = 120
Plugging t=36t = 36 and M(36)=15M(36) = 15 into M(t)=A(0.5)t/12M(t) = A \cdot (0.5)^{t/12} gives 15=A(0.5)315 = A \cdot (0.5)^3, which simplifies to 15=0.125A15 = 0.125A.

Anahtar Kavram

Interpreting and solving exponential decay functions with fractional exponents

Alternatif Yöntem

Instead of solving for dd first, recognize that 3636 hours is exactly 1.51.5 intervals of 2424 hours. Since the mass is multiplied by 0.250.25 (or 14\frac{1}{4}) every 2424 hours, after 3636 hours it will be multiplied by (14)1.5=(14)3/2=18(\frac{1}{4})^{1.5} = (\frac{1}{4})^{3/2} = \frac{1}{8} of its initial value. Therefore, 15=A18    A=12015 = A \cdot \frac{1}{8} \implies A = 120.
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