Math  /  Algebra

QuestionA sample of 700 g of an isotope decays to another isotope according to the function A(t)=700e0.033t\mathrm{A}(\mathrm{t})=700 e^{-0.033 \mathrm{t}}, where t is the time in years. (a) How much of the initial sample will be left in the sample after 25 years? (b) How long will it take the initial sample to decay to half of its original amount? (a) After 25 years, about 306.76 g of the sample will be left. (Round to the nearest hundredth as needed.) (b) After about \square years, the initial sample will decay to half of its original amount. (Round to the nearest hundredth as needed.)

Studdy Solution

STEP 1

What is this asking? We've got some radioactive material, 700 grams of it, and it's decaying over time.
We need to figure out how much is left after 25 years, and how long it takes for half of it to decay. Watch out! Don't mix up the initial amount with the amount remaining after decay.
Also, make sure to round correctly at the end!

STEP 2

1. Calculate Remaining Amount After 25 Years
2. Calculate Half-Life

STEP 3

Alright, let's **plug in** the time, t=25t = \textbf{25} years, into our decay function: A(t)=700e0.033tA(t) = 700e^{-0.033t}.
This function tells us how much of the isotope is left after a certain amount of time.

STEP 4

Substituting t=25t = \textbf{25}, we get A(25)=700e0.03325A(\textbf{25}) = 700e^{-0.033 \cdot \textbf{25}}.
Remember, that little dot means multiplication!

STEP 5

Now, let's **simplify** the exponent: 0.03325=-0.825-0.033 \cdot \textbf{25} = \textbf{-0.825}.
So, our equation becomes A(25)=700e-0.825A(\textbf{25}) = 700e^{\textbf{-0.825}}.

STEP 6

Using a calculator to evaluate e-0.825e^{\textbf{-0.825}}, we get approximately 0.4382\textbf{0.4382}.
So, A(25)=7000.4382A(\textbf{25}) = 700 \cdot \textbf{0.4382}.

STEP 7

Finally, **multiply** to find the remaining amount: 7000.4382306.74700 \cdot \textbf{0.4382} \approx \textbf{306.74} grams.

STEP 8

We want to find the time it takes for the sample to decay to half its original amount.
The **initial amount** is 700\textbf{700} grams, so half of that is 700/2=350\textbf{700} / 2 = \textbf{350} grams.

STEP 9

We'll **set** our decay function A(t)A(t) equal to 350\textbf{350} and **solve** for tt: 350=700e0.033t350 = 700e^{-0.033t}.

STEP 10

To **isolate** the exponential term, we'll **divide** both sides by 700\textbf{700}: 350/700=e0.033t350 / \textbf{700} = e^{-0.033t}, which simplifies to 0.5=e0.033t0.5 = e^{-0.033t}.
See how we divided to one?

STEP 11

Now, we'll take the **natural logarithm** (ln) of both sides to get rid of the exponential: ln(0.5)=ln(e0.033t)\ln(0.5) = \ln(e^{-0.033t}).
This simplifies to ln(0.5)=0.033t\ln(0.5) = -0.033t.
Remember, the natural log and the exponential function are inverse operations!

STEP 12

Using a calculator, we find that ln(0.5)-0.6931\ln(0.5) \approx \textbf{-0.6931}.
So, we have -0.6931=0.033t\textbf{-0.6931} = -0.033t.

STEP 13

Finally, we **divide** both sides by 0.033-0.033 to **solve** for tt: t=-0.6931/0.03320.99t = \textbf{-0.6931} / -0.033 \approx \textbf{20.99} years.

STEP 14

(a) After 25 years, approximately **306.74** g of the sample will be left. (b) After approximately **21.00** years, the initial sample will decay to half of its original amount.

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