Solved on Dec 03, 2023

Find the time it takes for a tranquilizer with a half-life of 50 hours to decay to $95\%$ of its original dosage using the exponential decay model $A=A_{0} e^{k t}$ .

STEP 1

Assumptions
1. The half-life of the tranquilizer is 50 hours
2. We need to find the time it takes for the drug to decay to 95% of the original dosage
3. We use the exponential decay model $A=A_{0} e^{k t}$ , where $A$ is the amount of drug at time $t$ , $A_{0}$ is the initial amount of drug, and $k$ is the decay constant

STEP 2

First, we need to find the decay constant $k$ . We can do this by using the formula for half-life, which is $t_{1/2} = \frac{\ln(2)}{k}$ .
$k = \frac{\ln(2)}{t_{1/2}}$

STEP 3

Now, plug in the given value for the half-life to calculate the decay constant.
$k = \frac{\ln(2)}{50}$

STEP 4

Calculate the decay constant.
$k = \frac{\ln(2)}{50} \approx 0.01386$

STEP 5

Now that we have the decay constant, we can use the exponential decay model to find the time it takes for the drug to decay to 95% of the original dosage. We set $A = 0.95A_{0}$ and solve for $t$ .
$0.95A_{0} = A_{0} e^{k t}$

STEP 6

Divide both sides of the equation by $A_{0}$ to isolate the exponential term.
$0.95 = e^{k t}$

STEP 7

Take the natural logarithm of both sides to solve for $t$ .
$\ln(0.95) = k t$

STEP 8

Substitute the value of $k$ into the equation.
$\ln(0.95) = 0.01386 t$

STEP 9

Solve for $t$ .
$t = \frac{\ln(0.95)}{0.01386}$

STEP 10

Calculate the time.
$t = \frac{\ln(0.95)}{0.01386} \approx 37.14 \, hours$
It will take approximately 37.14 hours for the drug to decay to 95% of the original dosage.

Was this helpful?

Find the time it takes for a tranquilizer with a half-life of 50 hours to decay to $95\%$ of its original dosage using the exponential decay model $A=A_{0} e^{k t}$ .

STEP 1

STEP 2

First, we need to find the decay constant $k$ . We can do this by using the formula for half-life, which is $t_{1/2} = \frac{\ln(2)}{k}$ .
$k = \frac{\ln(2)}{t_{1/2}}$

STEP 3

Now, plug in the given value for the half-life to calculate the decay constant.
$k = \frac{\ln(2)}{50}$

STEP 4

Calculate the decay constant.
$k = \frac{\ln(2)}{50} \approx 0.01386$

STEP 5

Now that we have the decay constant, we can use the exponential decay model to find the time it takes for the drug to decay to 95% of the original dosage. We set $A = 0.95A_{0}$ and solve for $t$ .
$0.95A_{0} = A_{0} e^{k t}$

STEP 6

Divide both sides of the equation by $A_{0}$ to isolate the exponential term.
$0.95 = e^{k t}$

STEP 7

Take the natural logarithm of both sides to solve for $t$ .
$\ln(0.95) = k t$

STEP 8

Substitute the value of $k$ into the equation.
$\ln(0.95) = 0.01386 t$

STEP 9

Solve for $t$ .
$t = \frac{\ln(0.95)}{0.01386}$

STEP 10

Calculate the time.
$t = \frac{\ln(0.95)}{0.01386} \approx 37.14 \, hours$
It will take approximately 37.14 hours for the drug to decay to 95% of the original dosage.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the time it takes for a tranquilizer with a half-life of 50 hours to decay to 95%95\%95% of its original dosage using the exponential decay model A=A0ektA=A_{0} e^{k t}A=A0​ekt.

STEP 1

STEP 2

First, we need to find the decay constant kkk. We can do this by using the formula for half-life, which is t1/2=ln⁡(2)kt_{1/2} = \frac{\ln(2)}{k}t1/2​=kln(2)​.k=ln⁡(2)t1/2k = \frac{\ln(2)}{t_{1/2}}k=t1/2​ln(2)​

STEP 3

Now, plug in the given value for the half-life to calculate the decay constant.k=ln⁡(2)50k = \frac{\ln(2)}{50}k=50ln(2)​

STEP 4

Calculate the decay constant.k=ln⁡(2)50≈0.01386k = \frac{\ln(2)}{50} \approx 0.01386k=50ln(2)​≈0.01386

STEP 5

Now that we have the decay constant, we can use the exponential decay model to find the time it takes for the drug to decay to 95% of the original dosage. We set A=0.95A0A = 0.95A_{0}A=0.95A0​ and solve for ttt.0.95A0=A0ekt0.95A_{0} = A_{0} e^{k t}0.95A0​=A0​ekt

STEP 6

Divide both sides of the equation by A0A_{0}A0​ to isolate the exponential term.0.95=ekt0.95 = e^{k t}0.95=ekt

STEP 7

Take the natural logarithm of both sides to solve for ttt.ln⁡(0.95)=kt\ln(0.95) = k tln(0.95)=kt

STEP 8

Substitute the value of kkk into the equation.ln⁡(0.95)=0.01386t\ln(0.95) = 0.01386 tln(0.95)=0.01386t

STEP 9

Solve for ttt.t=ln⁡(0.95)0.01386t = \frac{\ln(0.95)}{0.01386}t=0.01386ln(0.95)​

STEP 10

Calculate the time.t=ln⁡(0.95)0.01386≈37.14 hourst = \frac{\ln(0.95)}{0.01386} \approx 37.14 \, hourst=0.01386ln(0.95)​≈37.14hoursIt will take approximately 37.14 hours for the drug to decay to 95% of the original dosage.

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Find the time it takes for a tranquilizer with a half-life of 50 hours to decay to $95\%$ of its original dosage using the exponential decay model $A=A_{0} e^{k t}$ .

First, we need to find the decay constant $k$ . We can do this by using the formula for half-life, which is $t_{1/2} = \frac{\ln(2)}{k}$ .
$k = \frac{\ln(2)}{t_{1/2}}$

Now, plug in the given value for the half-life to calculate the decay constant.
$k = \frac{\ln(2)}{50}$

Calculate the decay constant.
$k = \frac{\ln(2)}{50} \approx 0.01386$

Now that we have the decay constant, we can use the exponential decay model to find the time it takes for the drug to decay to 95% of the original dosage. We set $A = 0.95A_{0}$ and solve for $t$ .
$0.95A_{0} = A_{0} e^{k t}$

Divide both sides of the equation by $A_{0}$ to isolate the exponential term.
$0.95 = e^{k t}$

Take the natural logarithm of both sides to solve for $t$ .
$\ln(0.95) = k t$

Substitute the value of $k$ into the equation.
$\ln(0.95) = 0.01386 t$

Solve for $t$ .
$t = \frac{\ln(0.95)}{0.01386}$

Calculate the time.
$t = \frac{\ln(0.95)}{0.01386} \approx 37.14 \, hours$
It will take approximately 37.14 hours for the drug to decay to 95% of the original dosage.