Math  /  Algebra

QuestionThe radioactive substance cesium-137 has a half-life of 30 years. The amount A(t)A(t) (in grams) of a sample of cesium-137 remaining after tt years is given by the following exponential function. A(t)=523(12)t30A(t)=523\left(\frac{1}{2}\right)^{\frac{t}{30}}
Find the initial amount in the sample and the amount remaining after 50 years. Round your answers to the nearest gram as necessary.
Initial amount: \square grams
Amount after 50 years: \square grams

Studdy Solution

STEP 1

What is this asking? We're given a formula that tells us how much cesium-137 is left after a certain number of years, and we need to figure out how much we started with and how much is left after 50 years. Watch out! Make sure to understand what a half-life means: it's the time it takes for *half* of the substance to decay.
Don't mix up the initial amount with the amount remaining after 50 years!

STEP 2

1. Find the initial amount.
2. Find the amount remaining after 50 years.

STEP 3

The **initial amount** is the amount of cesium-137 at the very beginning, when the time tt is **zero** years.
So, we need to **plug in** t=0t = 0 into our formula!

STEP 4

A(0)=523(12)030A(0) = 523 \cdot \left(\frac{1}{2}\right)^{\frac{0}{30}}

STEP 5

Any number raised to the power of zero is **one**, except for zero itself, but we don't have to worry about that here!
So, (12)030=(12)0=1\left(\frac{1}{2}\right)^{\frac{0}{30}} = \left(\frac{1}{2}\right)^0 = 1.

STEP 6

Now, we have: A(0)=5231=523A(0) = 523 \cdot 1 = 523 So the **initial amount** is 523\textbf{523} grams.

STEP 7

Now, we want to find the amount remaining after t=50t = \textbf{50} years.
Let's **plug in** t=50t = 50 into our formula:

STEP 8

A(50)=523(12)5030A(50) = 523 \cdot \left(\frac{1}{2}\right)^{\frac{50}{30}}

STEP 9

We can simplify the exponent 5030\frac{50}{30} to 53\frac{5}{3}.

STEP 10

A(50)=523(12)53A(50) = 523 \cdot \left(\frac{1}{2}\right)^{\frac{5}{3}}

STEP 11

Now, we can use a calculator to evaluate (12)53\left(\frac{1}{2}\right)^{\frac{5}{3}}.
This is approximately **0.315**.

STEP 12

A(50)5230.315164.745A(50) \approx 523 \cdot 0.315 \approx 164.745

STEP 13

Rounding to the nearest gram, we get **165** grams.

STEP 14

Initial amount: **523** grams Amount after 50 years: **165** grams

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