Math

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

Studdy Solution

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=A0ektA=A_{0} e^{k t}, where AA is the amount of drug at time tt, A0A_{0} is the initial amount of drug, and kk is the decay constant

STEP 2

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

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}

STEP 4

Calculate the decay constant.
k=ln(2)500.01386k = \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=0.95A0A = 0.95A_{0} and solve for tt.
0.95A0=A0ekt0.95A_{0} = A_{0} e^{k t}

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

Calculate the time.
t=ln(0.95)0.0138637.14hourst = \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.

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