## anonymous 3 years ago When a patient is given a certain quantity Q0 of medication, in grams, the liver and kidneys eliminate about 40% of the drug from the bloodstream each hour, so that only 60% of the drug will remain in the system after each hour. We let Q(t) be the quantity of drug available in the body at any time t, in hours. (a) If Q0 = 250 mg, find Q(1), Q(2), and Q(3). (b) Find a formula for Q(t) as an exponential function of t, for t ≥ 0. (c) At what time t does 75 mg of the drug remain?

1. anonymous

I don't even have a clue as how to start or even what to google or look-up in my book for help

2. anonymous

I got the answer for A, that was pretty straight forward

3. anonymous

I have Q(1)=150, Q(2)=90, Q(3)=54

4. zepdrix

Hmm yah those looks correct. Part B) now huh? Hmmmm.

5. anonymous

would it be like Q(t)=250-.6^t?

6. anonymous

nope that doesnt work if i plug numbers in.. just a thought

7. anonymous

oh its just 250*.6^t

8. zepdrix

Yay good job! C:

9. anonymous

I figured that out by trail an error. I guess thats the way to do it! Lol thanks for the back up again

10. anonymous

thanks! gave you a medal back

11. zepdrix

Able to figure out part C ok? :)

12. anonymous

set it equal to 54 and solve for t, right?

13. anonymous

i meant 75

14. zepdrix

Yah c: cool.

15. anonymous

how would i get the t out of the exponent? I know to just divide by 250 on both sides

16. zepdrix

We have to use that nasty logarithm function to get it out of the exponent position! :O

17. anonymous

then i'm left with $\frac{ 75 }{ 250 }=.6^{t}$

18. anonymous

ohh ew. hows that go again?

19. zepdrix

$\large .3=.6^t$If we take the natural log of both sides,$\large \ln .3=\ln\left(.6^t\right)$

20. zepdrix

Then we need to remember a handy rule of logs,$\huge \log(a^{\color{orangered}{b}})=\color{orangered}{b} \log(a)$

21. anonymous

so ln.3=tln.6?

22. anonymous

then just plug the two logs into my calculator and then divide over to solve for t?

23. zepdrix

Yep looks good \c:/ Mr Calculator has to finish it up for us!

24. anonymous

awesome. thanks as always!