Convert Q = 973.6^(2.1t) to form ae^(kt); half-life; solve

[kristenf]

So I have to admit that even with seeing a one on one tutor, reviewing all of my past homework assignments, and so on - I made a 37 on my first precal test. I don't know HOW this happened, but I think I have a problem that I think I know how to do a problem, but I really don't.

Whatever i'm doing does not seem to be working..

I need help on these problems. A step by step analysis would be great, or maybe even just help on HOW to go about solving these problems. It's not that I can't do the work - I just don't know where to begin. I do try to pay attention in class but I am usually also busy jotting down notes so I sometimes don't remember the material we covered in class.

Here are some problems I am having trouble with:

(1) Convert to the form Q = ae ^kt

Q= 973.6 ^ 2.1t


*I dont know how to convert. I've looked in the book but cannot find a good example.
If I am shown the steps I can solve this

(2) Find the half life of the substance:

decays at a rate of 3.406% per day


(3) Solve the equations (give an exact solution if there is one)

10e ^3t - e = 2e ^ 3t


(4) What is the value (if any) of the following?

a) 10 ^ -x as x ---> oo (that is a symbol)

b) lnx as x ----> 0+


(5) find the domain of the function:

f(x) = ln(lnx)

 

[mmm4444bot]

Re: I don't need answers I need help on HOW to solve/steps.. =[

... I made a 37 on my first precal test. I don't know HOW this happened ...

Hello Kristen:

Perhaps you have not completed enough exercises to meet your own expectations. Do more exercises to get more experience.

(1) Convert to the form Q = a*e^(k*t)

Q= (973.6)^(2.1t)

Use the given equation to find Q when t = 0 and t = 1.

Then substitute t = 0 and the corresponding value for Q into the desired form to get an equation with a and k.

Do the same thing with t = 1 and the crresponding value for Q to get a second equation with a and k.

Solve this system.

EG: Q = (5)(4.096)^(t/3)

Q = 5 when t = 0.

Q = 8 when t = 1.

Substitue Q and t values into the form Q = a*e^(k*t)

5 = a*e^(0*k)

8 = a*e^(1*k)

The first equation simplifies to a = 5.

Therefore, the second equation becomes

8 = 5*e^k

e^k = 8/5

k = ln(8/5)

k = 0.47

Q = 5*e^(0.47*t)

(2) Find the half life of the substance:

decays at a rate of 3.406% per day

I'll do a similar example; I don't want it to be the same problem with different numbers, so I'll use a different time unit (weeks).

EG: decays 9.8% per week; find the half-life in weeks.

Determine what percent remains after one week.

100 - 9.8 = 90.2

90.2% remains. How many times do we need to multiply by 0.902 to get to 50% (i.e., half)?

(0.902)^w = 0.50

w * ln(0.902) = ln(0.50)

w = ln(0.50)/ln(0.902)

w = 6.72

The half-life is about 6.7 weeks.

(3) Solve the equations (give an exact solution if there is one)

10e ^3t - e = 2e ^ 3t

Remember that e is a constant; you are solving for t.

Combine the like-terms on one side, and move the constant e to the other side. Divide by 8.

Solve by taking the natural log of both sides. You'll end up with a factor of ln(e/8), but e is just a number. So put e/8 into your calculator's natural log function by entering 2.71828/8.

(4) What is the value (if any) of the following?

a) 10 ^ -x as x ---> oo (that is a symbol)

Yes, oo is a symbol. It's a special symbol. It does not represent a number; it represents an idea. That idea is infinity. When you do math with infinity, you need to use analysis instead of arithmetic because infinity is not a number.

Remember that a negative sign in front of the exponent creates a reciprocal.

10^(-x) = 1/10^x

Think about the value of 10^x as x becomes very large.

Think about what happens to the value of a ratio when the numerator is fixed and the denominator becomes either tiny or huge.

b) lnx as x ----> 0+

Become familiar with the graph of the natual log function because this function arises in many different applications. Follow the curve of this graph as you approach the y-axis (i.e., when the value of x is approaching zero from the right). Where does the curve of ln(x) take you in this case? This is what part (b) is asking for.

(5) find the domain of the function:

f(x) = ln(lnx)

(Did I just write that the natural log function keeps coming up?)

We know that ln(x) is only defined for positive values of x. In other words, we cannot put zero or any negative real number into the natural log function.

This is another way of saying that the domain of ln(x) is x>0.

Well, this exercise has a composite function. Instead of putting x into the natural log function, we're putting ln(x) into the natural log function.

(We are composing ln(x) with itself.)

So, ln(x) must be greater than zero in order for ln[ln(x)] to make sense. What values of x cause ln(x)>0?

This set of values is the domain of ln[ln(x)].

Again, if you're familiar with the basic features of the graph of ln(x), then questions like "when is ln(x)>0?" become easy to answer.

Cheers,

~ Mark

 

[Deleted member 4993]

Re: I don't need answers I need help on HOW to solve/steps.. =[

So I have to admit that even with seeing a one on one tutor, reviewing all of my past homework assignments, and so on - I made a 37 on my first precal test. I don't know HOW this happened, but I think I have a problem that I think I know how to do a problem, but I really don't.

Whatever i'm doing does not seem to be working..

I need help on these problems. A step by step analysis would be great, or maybe even just help on HOW to go about solving these problems. It's not that I can't do the work - I just don't know where to begin. I do try to pay attention in class but I am usually also busy jotting down notes so I sometimes don't remember the material we covered in class.

Here are some problems I am having trouble with:

(1) Convert to the form Q = ae ^kt

Brush up on laws of logarithm and exponentials - you'll use it often in this class and beyond.....

\(\displaystyle Q \, = \, 973.6 ^ {2.1t}\)

\(\displaystyle Q \, = [(973.6)^{2.1}]^t\)

\(\displaystyle Q \, = [1886249.714]^t\)

\(\displaystyle Q \, = [e^{ln(1886249.714)}]^t\)

\(\displaystyle Q \, = [e^{14.45}]^t\)

Now finish it....