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When a person has a cavity filled, the dentist typically administers a local anesthetic. After leaving the dentist's office, one's mouth often feels numb for several hours. If a shot of anesthesia is injected into the bloodstream at the time of the procedure (t=0), and the amount of anesthesia still in the bloodstream t hours after the initial injection is given by A=A0e^-0.5t, in how many hours will only 10% of the original anesthetic still be in the bloodstream?
we are solving
.1 = 1(e^(-.5t)
take ln of both sides
ln .1 = ln (e^(-.5t))
ln .1 = -.5t (1)
t = ln .1/-.5 = 4.61 hrs
= appr. 4 hours and 36 minutes
To find the number of hours it will take for only 10% of the original anesthetic to be in the bloodstream, we can set up an equation using the given formula A = A0e^(-0.5t).
Step 1: Substitute the given values into the equation.
We know that A is the amount of anesthesia in the bloodstream after t hours, and A0 is the initial amount of anesthesia injected into the bloodstream at t=0.
Let's call the initial amount of anesthesia A0 and the time it takes for only 10% to be in the bloodstream t_10.
So, we have:
A = 0.1A0 (10% of the initial amount)
Now we can substitute these values into the given equation:
0.1A0 = A0e^(-0.5t_10)
Step 2: Solve for t_10.
Let's divide both sides of the equation by A0 to isolate the exponential term:
0.1 = e^(-0.5t_10)
Step 3: Take the natural logarithm of both sides.
To solve for t_10, we need to take the natural logarithm (ln) of both sides of the equation:
ln(0.1) = ln(e^(-0.5t_10))
Since ln(e^(x)) = x, we can simplify the equation:
ln(0.1) = -0.5t_10
Step 4: Solve for t_10.
To isolate t_10, divide both sides of the equation by -0.5:
t_10 = ln(0.1) / -0.5
Using a calculator, we can find the value of t_10:
t_10 ≈ (ln(0.1)) / -0.5 ≈ 3.4657
Therefore, it will take approximately 3.4657 hours for only 10% of the original anesthetic to be in the bloodstream.