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Instructions: Create the equation of the form [tex]y = a \cdot b^x[/tex] for the exponential function described in each real-world problem. Then, use the equation to answer the question.

If there are 180 grams of radioactive material with a half-life (decrease by half or [tex]50\%[/tex]) of 1 hour, how much of the radioactive material will be left after 3 hours?

1. What number will you fill in for [tex]x[/tex] to solve the equation?
[tex]x = \square[/tex]

2. What will be the value of [tex]y[/tex] after 3 hours?
[tex]y = \square[/tex]


Sagot :

Sure! Let's solve this step-by-step.

### Problem Analysis
We are given:
- An initial amount of radioactive material: 180 grams
- A half-life of 1 hour, meaning the material decreases by half every hour
- We need to determine the amount of radioactive material left after 3 hours.

### Step-by-Step Solution

1. Identify the parameters for the exponential decay formula.

The general form of the exponential decay formula is:
[tex]\[ y = a \left( \frac{1}{2} \right)^{\frac{x}{b}} \][/tex]

Where:
- [tex]\( y \)[/tex] is the remaining amount of material after time [tex]\( x \)[/tex]
- [tex]\( a \)[/tex] is the initial amount of material
- [tex]\( \frac{1}{2} \)[/tex] represents the half-life decay factor
- [tex]\( x \)[/tex] is the elapsed time
- [tex]\( b \)[/tex] is the half-life period

2. Substitute the given values into the formula.
- Initial amount [tex]\( a = 180 \)[/tex] grams
- Half-life period [tex]\( b = 1 \)[/tex] hour
- Elapsed time [tex]\( x = 3 \)[/tex] hours

The formula becomes:
[tex]\[ y = 180 \left( \frac{1}{2} \right)^{\frac{3}{1}} \][/tex]

3. Simplify the exponent.
[tex]\[ y = 180 \left( \frac{1}{2} \right)^3 \][/tex]

4. Calculate the decay factor raised to the power.
[tex]\[ \left( \frac{1}{2} \right)^3 = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \][/tex]

5. Now substitute this back into the equation.
[tex]\[ y = 180 \times \frac{1}{8} \][/tex]

6. Perform the multiplication.
[tex]\[ y = 180 \times 0.125 = 22.5 \][/tex]

### Answer
After 3 hours, there will be [tex]\( 22.5 \)[/tex] grams of the radioactive material left.

To summarize:
- The value for [tex]\( x \)[/tex], the elapsed time, is [tex]\( 3 \)[/tex] hours, filling in the first blank.
- The equation we derived is [tex]\( y = 180 \left( \frac{1}{2} \right)^3 \)[/tex].
- Substituting the numeric values yields [tex]\( y = 22.5 \)[/tex].

So, you should fill in:
1. [tex]\( x = 3 \)[/tex]
2. [tex]\( y = 22.5 \)[/tex]
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