The amount remaining of a radioactive substance after t hours is given by

A(t) = 100ekt. After 12 hours, the initial amount has decreased by 4%.

How much remains after 48 hours?

What is the half-life of the substance?

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for the half life, set the amount = 50 and solve for t

The amout remaining of a radioactive substance after t hours id given by A(t)=100e^kt. After12hours, the initial amount had decreased by 7%

To find the amount remaining after 48 hours, we need to substitute t = 48 into the equation A(t) = 100e^kt and solve for A(48).

From the given information, we know that after 12 hours, the initial amount has decreased by 4%. This means that the remaining amount is 100% - 4% = 96% of the initial amount.

Substituting t = 12 into the equation A(t) = 100e^kt and equating it to 96% of the initial amount, we can solve for k.

96% of A(0) = A(12) = 100e^12k
0.96A(0) = 100e^12k
e^12k = 0.96

Taking the natural logarithm of both sides gives:
12k = ln(0.96)
k = ln(0.96)/12

Now we can substitute k into the equation A(48) = 100e^kt and calculate the remaining amount after 48 hours.

A(48) = 100e^(48ln(0.96)/12)
A(48) ≈ 100e^0.0391
A(48) ≈ 100 * 1.0398
A(48) ≈ 103.98

Therefore, approximately 103.98 units of the substance remain after 48 hours.

To find the half-life of the substance, we can set A(t) = 100e^kt to half of its initial amount, A(0)/2 = 100/2 = 50.

50 = 100e^kt
e^kt = 0.5

Taking the natural logarithm of both sides gives:
kt = ln(0.5)

The half-life, t_half, is the time it takes for the remaining amount to decrease to half its initial value. Thus, we can solve for t_half:

t_half = ln(0.5)/k
t_half = ln(0.5)/(ln(0.96)/12)

Evaluating this expression gives:
t_half ≈ -ln(0.5) * 12/ln(0.96)
t_half ≈ 18.201 hours

Therefore, the half-life of the substance is approximately 18.201 hours.

To find the amount remaining after 48 hours, you need to evaluate the equation A(t) = 100ekt at t = 48.

Step 1: Given that the initial amount has decreased by 4%, we can find the decay constant, k.

Let's assume the initial amount A(0) = A0 and the decreased amount after 12 hours is A(12) = 0.96A0 (4% decrease).

Using the formula A(t) = A0e^kt, we can substitute the values:

0.96A0 = A0e^(12k)

Divide both sides by A0:

0.96 = e^(12k)

Take the natural logarithm of both sides:

ln(0.96) = 12k

Divide by 12:

k = ln(0.96) / 12

Step 2: Substitute the value of k into the equation A(t) = 100ekt and evaluate it at t = 48:

A(48) = 100e^(48k)

Replace the value of k:

A(48) = 100e^(48 * ln(0.96) / 12)

Use a calculator to compute this value.

The half-life of a radioactive substance is the time it takes for the initial amount to decrease by half.

The half-life can be found by setting the remaining amount A(t) to half of the initial amount, A0:

0.5A0 = A0e^(kt)

Divide both sides by A0:

0.5 = e^(kt)

Take the natural logarithm of both sides:

ln(0.5) = kt

Solve for t:

t = ln(0.5) / k

Substitute the value of k we found earlier and calculate the half-life using a calculator.