The breakdown of sample of chemical compound is represented by the function `p(t)=300((1)/(2))^(t)`, where p(t) represents the number of milligrams of the substance, and t represents the time , in years. If `t=0` represents the year 2015, what will be the first year in which the amount of the substance remaining falls to cless than 5 milligrams?

Answer

To solve the problem step by step, we need to determine when the function \( p(t) = 300 \left( \frac{1}{2} \right)^t \) falls below 5 milligrams.

### Step 1: Set up the inequality
We start by setting up the inequality based on the problem statement:
\[ p(t) < 5 \]
Substituting the function into the inequality gives:
\[ 300 \left( \frac{1}{2} \right)^t < 5 \]

### Step 2: Isolate the exponential term
Next, we divide both sides of the inequality by 300 to isolate the exponential term:
\[ \left( \frac{1}{2} \right)^t < \frac{5}{300} \]
This simplifies to:
\[ \left( \frac{1}{2} \right)^t < \frac{1}{60} \]

### Step 3: Rewrite the inequality
We can rewrite the inequality in terms of powers of 2. Since \( \left( \frac{1}{2} \right)^t = \frac{1}{2^t} \), we can express the inequality as:
\[ \frac{1}{2^t} < \frac{1}{60} \]
This implies:
\[ 2^t > 60 \]

### Step 4: Find the smallest integer \( t \)
Now, we need to find the smallest integer \( t \) such that \( 2^t > 60 \). We can calculate the powers of 2:
- \( 2^5 = 32 \)
- \( 2^6 = 64 \)

Since \( 2^6 = 64 \) is the first power greater than 60, we have:
\[ t = 6 \]

### Step 5: Determine the corresponding year
Since \( t = 0 \) corresponds to the year 2015, we find the year when \( t = 6 \):
\[ 2015 + 6 = 2021 \]

### Conclusion
The first year in which the amount of the substance remaining falls to less than 5 milligrams is **2021**.

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