# exponential word problem

• Jan 19th 2010, 08:32 AM
Tweety
exponential word problem
• Jan 19th 2010, 08:49 AM
earboth
Quote:

Originally Posted by Tweety

to a)

Plug in D = 10 and t = 5. Use a calculator. I've got $x \approx 5.353\ \rm{mg}$

to b)

Now D = 10 + 5.353 and t = 1. Use a calculator. I've got $x \approx 13.54897\ \rm{mg}$

to c)

Solve the equation for t:

$3 = 15.353\cdot e^{-\frac18 t}~\implies~\dfrac{3}{15.353} = e^{-\frac18 t}$

Use logarithm:

$\ln\left(\dfrac{3}{15.353} \right) = -\frac18 t$

I'll leave the rest for you.
Spoiler:
I've got $t \approx 13.06\ h$, that means nearly 13 hours and 4 minutes
• Jan 19th 2010, 08:56 AM
Tweety
Thank you,

but D= the amount of dose given, so why have you added the answer to part 'a' , which is the amount of the drug in the bloodstream?

Cause I thought to put D = 20,
• Jan 19th 2010, 09:04 AM
earboth
Quote:

Originally Posted by Tweety
Thank you,

but D= the amount of dose given, so why have you added the answer to part 'a' , which is the amount of the drug in the bloodstream?

Cause I thought to put D = 20,

The text of b) reads: A second dose is given after 5 hours.

From a) I know the amount of the drug after 5 hours and to this amount a second dose is added.