FlashFlashRevolution [High School - Math] Exponential word problems
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 02-26-2008, 08:11 PM #1 21death12 FFR Player     Join Date: Jan 2007 Posts: 5 [High School - Math] Exponential word problems A cough medicine is expected to kill one half the cough germs each time the medicine is taken. Between uses the number of germs increases by one-third. If the medicine can only be taken every 4 hours, and complete relief is obtained when the germs have been reduced to one-tenth the original amount, how many times must the medicine be taken to obtain complete relief? I'm more interested in the equation to figure it out than the answer since I already have that. (6) The equation I got was 10%=100%(0.833)^n/4 I did -(1/2)+(1/3) to get the decay rate though I don't think thats right. Everytime i try to find n it doesnt come close to 6 so can someone explain if I have the right equation or explain how to get another one? Thanks
 02-26-2008, 09:18 PM #2 Goldenwind FFR Simfile Author FFR Simfile Author     FFR Veteran Join Date: Jan 2006 Location: Canada Age: 25 Posts: 761 Re: [High School - Math] Exponential word problems At t=0, you have 100% germs Also at t=0, you take one dose, reducing it to 50% At t=4(hours), it has risen by 1/3rd, aka to 66.7% Also at t=4, you take your second dose, reducing it to 33.3% At t=8(hours), it has risen by 1/3rd, aka to 44.4% Also at t=8, you take your third dose, reducing it to 22.2% Immediately after the uses: t=0, germs = 50% t=4, germs = 33.3% t=8, germs = 22.2% To help you see the equation: Start at 100% Divide by 2 Multiply by (1+1/3) = 4/3 So, each time you're multiplying by a factor of (1/2)*(4/3) = 4/6 = 2/3, of the previous value. So, #germs = 50%*(2/3)^(t/4) When t=0, #germs = 50%*(2/3)^(0/4) #germs = 50%*(2/3)^0 #germs = 50%*1 #germs = 50% When t=4, #germs = 50%*(2/3)^(4/4) #germs = 50%*(2/3)^1 #germs = 50%*(2/3) #germs = 33.3% When t=8, #germs = 50%*(2/3)^(8/4) #germs = 50%*(2/3)^2 #germs = 50%*(2/3)*(2/3) #germs = 33.3%*(2/3) #germs = 22.2% And so on. Number of doses it'll take for you to get below 10%? Make a new variable k, representing the number of doses. When t = 0, k = 1 When t = 4, k = 2 When t = 8, k = 3 When t = 12, k = 4 And so on. (Assuming you took your first dose immediately) So, just from observing it, k = (t/4)+1 Or, t = 4(k-1) Now sub this into our formula: #germs = 50%*(2/3)^(t/4) 10% = 50%*(2/3)^(4(k-1)/4) 10% = 50%*(2/3)^(k-1) 0.2% = (2/3)^(k-1) ln(0.2) = ln((2/3)^(k-1)) ln(0.2) = (k-1)ln(2/3) k = (ln(0.2) / ln(2/3)) + 1 k = (-1.61 / -0.41) + 1 k = (-1.61 / -0.41) + 1 k = 3.97 + 1 k = 4.97 It takes 4.97 doses to reduce it to 10%. However, we can only take integer doses, and the closest integer that is above 4, is 5, so... k ~= 5 For 5 doses, this will bring it a bit under 10%, but it's the first integer that will do so. If you took your first dose at t=4, then t=0, 100% t=4, grown by 1/3rd (133%), cut in half (66.7%) t=8, grown by 1/3rd (88.9%), cut in half (44.4%) So, using the same logic: #germs = 100%*(2/3)^(t/4) #germs = 100%*(2/3)^(t/4) #germs = 100%*(2/3)^(0/4) #germs = 100% #germs = 100%*(2/3)^(t/4) #germs = 100%*(2/3)^(4/4) #germs = 66.7% Etc. When t = 0, k = 0 When t = 4, k = 1 When t = 8, k = 2 When t = 12, k = 3 And so on. So, just from observing it, k = (t/4) Or, t = 4k Now sub this into our formula: #germs = 100%*(2/3)^(t/4) 10% = 100%*(2/3)^(4k/4) 10% = 100%*(2/3)^k 0.1% = (2/3)^k ln(0.1) = ln((2/3)^k) ln(0.1) = k*ln(2/3) k = (ln(0.1) / ln(2/3)) k = (-2.30 / -0.41) k = (-2.30 / -0.41) k = 5.68 k ~= 6 If you take one immediately, it'll take 5 doses. Else, 6. A general rule for growth functions: A = A_0 * r^(t/n) Where A_0 is the starting value Where r is the rate of growth Where t is time Where n is increments of time (For example, here n = 4, as the growth was applied ever 4 hours, not ever 1) And where A = the value of something, after having started at A_0, and grown for time t. A special case of this, which you may see later, is y = y_0 * e^(kt), or ln(y/y_0) = kt (Since ln(e) = 1) __________________ They say there's a sun in the sky... Last edited by Goldenwind; 02-26-2008 at 10:09 PM..
 02-26-2008, 10:35 PM #3 21death12 FFR Player     Join Date: Jan 2007 Posts: 5 Re: [High School - Math] Exponential word problems Thanks GoldenWind We haven't done natual logarithms yet we're only using log with base 10 right now but I can see how to get equation now thanks again
 02-26-2008, 10:36 PM #4 -Moo- FFR Player     Join Date: Jan 2007 Location: The Desert, Arizona Posts: 1,199 Re: [High School - Math] Exponential word problems que? __________________ touch my pokeballz
02-27-2008, 12:32 AM   #5
Goldenwind
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Re: [High School - Math] Exponential word problems

Quote:
 Originally Posted by 21death12 Thanks GoldenWind We haven't done natual logarithms yet we're only using log with base 10 right now but I can see how to get equation now thanks again
ln(x) = log[base e](x)
e is a special number (Like pi), which is approx 2.71.

log[base10](10) = 1
log[base e](e) = ln(e) = 1

I use ln() from habit. Where I used ln() above, log() also works.

And the one rule you need to know is the log power rule:

log(a^b) = b*log(a)

Good luck
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