[High School - Math] Exponential word problems

[21death12]

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

[Goldenwind]

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)

[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

[-Moo-]

que?

[Goldenwind]

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