still cant do logarithms???

Question

During surgery, a patient must have at least 40 mg of an antibiotic in his system. The amount of anitbiotic present in (k) hours after administration of 100 mg of this antibiotic is given by P(k)=100(.508)^k. After how many hours to nearest tenth will the nurse have to give another dose to keep the level of antibiotic high enough?

Answer 1

40 = 100(0.508)^k
0.4 = 0.508^k
k = log(base 0.508) 0.4
k = ln 0.4 / ln 0.508
k = 1.3529 hours

Since the level will have dropped below 40 mg by 1.4 hours, the nurse must give another dose after 1.3 hours.

Answer 2

It's really simple. You already know that the patient needs at least 40 mg of antibiotic. You are already given the formula for the amount of antibiotic in the patient's system after k number of hours. You just need to plug in the variables in the equation.

P(k)=100*(.508)^k
40=100*(.508)^k
40/100 = 0.508^k


Now, in this case, how would you solve for k? Well, remember that one of the properties of logs is the following:
logb(m^n) = n * logb(m), where b = the base number.. in this case, we'll just use b = 10. So that would be:
log10(m^n) = n * log10(m) ... in standard calculators the LOG button normally defaults to base 10, so you just need to punch in LOG.


If we apply the property to the original equation, then we get:

log10(40/100) = log10(0.508^k)
log10(40/100) = k * log10(0.508)
k = 1.3529 hrs


In a more readable format, that would be 1 hr 21 mins 10 secs. To get this, is very simple. Originally, k is already in the hrs unit. So, we know that it is 1 hour plus _x_ minutes and _y_ seconds.

To calculate the minutes and seconds we just need to multiply 60 minutes * 0.3529 and we get 21.174 minutes. The reason for this is because we know that we have an additional 0.3529 hrs from calculating k. With 1 hrs = 60 mins, we can do the following conversion:

(60 mins / 1 hrs) * (0.3529 hrs) = 21.174 minutes, the hrs unit cancels out.


To calculate the seconds, we just need to multiply 60 seconds * .174, and we get approximately 10 seconds.

Answer 3

40 = 100(.508)^k
k = ln(40/100)/ln.508 = 1.4 hrs

Answer 4

P(k) = 100(.508)^k
40 = 100(0.508)^k
0.4 = (0.508)^k
log(0.4) 0.4 = log(0.4) (0.508)^k
1 = k*log(0.4) (0.508)
1/log(0.4) (0.508) = k
k = 1.35291028

so after approx 1.4 hours the nurse will have to give the patient another dose of antibotic.

Answer 5

100(.508)^k = 40
(.508)^k = .4
ln (.508)^k = ln .4
k ln .508 = ln .4
k = ln .4 / ln .508 = 1.4 hours