Growth/Decay Question: f(x) = 100(1/2)^(t/2): find time

[HorseWhisperz]

Okay, so here was my problem:

If the amount of drug remaining in the body after t hours is given by f(x)=100(1/2)[sup:1q67kioq]t/2[/sup:1q67kioq], then calculate the number of hours it would take for the initial 100 mg to become 35 mg.

Round to two decimal places.

Here's what I did... maybe someone can point out where I went wrong.

A(t)= A[sub:1q67kioq]0[/sub:1q67kioq]e[sup:1q67kioq]kt[/sup:1q67kioq]
A(t)=35
A[sub:1q67kioq]0[/sub:1q67kioq]=100
k=0.5
t=unknown

e[sup:1q67kioq]0.5t[/sup:1q67kioq]= 35/100
ln e[sup:1q67kioq]0.5t[/sup:1q67kioq]= ln(35/100)
0.5t=ln(35/100)
t= (ln 35/100)/0.5

but then I get a negative t.... which is not possible, correct?

 

[mmm4444bot]

… given by f(x) = 100 * (1/2)^(t/2) …

… Here's what I did …

A(t) = A_0 * e^(kt)

Whoops. You're using the wrong model.

In this exercise, they've chosen the base to be 1/2, not e. And, they've given you the value of "k". It's 1/2, also.

If you'd like to learn something about notation, they've chosen a name for the given function. They chose f.

And, I'm sure that they did not switch names on the independent variable from x on the left-hand side to t on the right-hand side, so that's a typographical error.

Solve the following for t.

f(t) = 35

In other words,

100 * (1/2)^(t/2) = 35

You'll get an expression for t that contains a base-1/2 logarithm. You can then use the change-of-base formula to obtain a decimal approximation using a scientific calculator.

\(\displaystyle 2 \cdot \frac{ln(\frac{7}{20})}{ln(\frac{1}{2})}\)

Let me know if I typed anything that you do not understand, or if you get stuck in the solution process. Thanks for showing your work.

 

[HorseWhisperz]

Re: Growth/Decay Question

I understand that I used the wrong formula, but I'm just so confused what to do. I haven't done logarithms and growth/decay in about three years.

 

[mmm4444bot]

… I'm just so confused what to do. I haven't done logarithms and growth/decay in about three years …

Well, you correctly solved the equation A(t) = 35. The steps are the same for solving f(t) = 35.

The only difference is that you'll be taking the "log-base-one-half" of each side of the equation, instead of taking the natural logarithm (log-base-e) of each side.

After you get the expression for t, the change-of-base formula lets you express the log-base-1/2 part in terms of either log-base-10 (LOG) or log-base-e (the natural logarithm, ln). Scientific calculators have buttons for these two bases, so that you can evaluate your expression for t.

Since you're "so confused", I'm going to be verbose in this post.

To use natural lograrithms, the formula is: log_b(a) = ln(a)/ln(b)

In words, "The log-base b of a is equal to the natural log of a divided by the natural log of b."

To use base-10 logarithms, the formula is: log_b(a) = LOG(a)/LOG(b)

In words, "The log-base-b of a is equal to the log-base-10 of a divided by the log-base-10 of b."

EXAMPLE: What power does 2 need to be raised to in order to equal 1/16th? In other words, what is x such that

2^x = 1/16

We can express this unknown exponent using logarithmic notation.

x = log_2(1/16)

To evaluate this number, I'll use the natural logarithm change-of-base formula.

log_2(1/16) = ln(1/16)/ln(2).

Dividing ln(1/16) by ln(2) on a calculator yields -4.

2^(-4) = 1/16

EXAMPLE: Change log-base-4 of 17 into log-base ten to evaluate a decimal approximation to four places. (This will answer the question to what exponent 4 needs to be raised in order to equal 17.)

log_4(17)

The base b = 4

The value of the power a = 17

log_b(a) = LOG(a)/LOG(b)

log_4(17) = LOG(17)/LOG(4)

"Log-base-4 of 17 is equal to log-base-10 of 17 divided by log-base-10 of 4."

Use a calculator's LOG button to evaluate this division.

You'll get the same value by using the other formula and the calculator's ln button.

log_4(17) = ln(17)/ln(4)

And, for review, remember that logarithms are exponents.

log_b(a) = x

b^x = a

If I tell you that the value of log_4(17) is equal to 2.0437 (rounded to four decimal places), then I'm telling you that some exponent is 2.0437.

Specifically, it's the exponent needed to get the power of four that equals 17.

4^(2.0437) = 17

This equation shows the relationship between the numbers 4, 17, and 2.0437 expressed exponentially. I typed it so that you can clearly see that the base involved is b = 4 and that the value of the power involved is a = 17.

We can express the same relationship between these numbers logarithmically.

log_4(17) = 2.0437

In your work with A(t), you wound up with:

t = log_e(0.35)/0.5

(Here, I've typed the natural logarithm as log-base-e, instead of using the ln notation; same thing)

"t equals the log-base-e of 0.35 divided by one-half."

With f(t) in your exercise, the base is 0.5, instead of e. Since the steps are the same, you're going to wind up with log_0.5 instead of log_e:

t = log_0.5(0.35)/0.5

t = 2 * log_0.5(0.35)

(Dividing by 0.5 is the same as multiplying by 2, right?)

Now, to evaluate this expression t, you need to get a decimal approximation for "log-base-one-half of 0.35", and double it.

Use one of the change-of-base formulas above to evaluate log_0.5(0.35) in terms of either ln or LOG on a calculator.

Rounded to the nearest hundreth of a second, I get 3.03 seconds for the body to reduce the amount of whatever stuff they're modeling to 35 mg.

Checking: f(3.03) = 35 ?

100 * (1/2)^(3.03/2) = 34.99

It checks.

I realize that I wrote a lot; please feel free to ask questions.