Precal Blog

Today in class, we learned a lot of new information about logarithms.  For those of you that are still a little bit uneasy about it, I will try my best and explain it.

First, we finished up talking about number 2 on activity 2.4.    In the first parts of activity 2.4, we were looking at the relationship between the exponent and power number lines.  We replicatied these number lines on our papers so we could easily measure from point to point.  Our conclusions were:

• For everychange of 1 for exponent, there is a multiplication of 10 for the power.
• For every change of 2 for exponent, there is a mulitplication of 100 for the power.

Knowing this simple knowledge of those relationships allowed us to dig even deeper into the material.  Our next table dealt with finding the exponent values when there was a change of *2 in the power value.  For this we measured from 1 to 2 on our power line and the exponent line and compared the ratio.  1 to 2 on the power line meaured about 18 mm.  1 to 2 on the exponent line measured about 59 mm.  We then concluded the ratio was about .301.  This means that every time you multiply 2 on the power line, you add .301 on the exponent line.  I tried my best to show that visually below:

Reversearp gave us a few examples of applying logarithms.  For example:

log64

log64 = log(2^6)

= 6(.301)

We then got the equation log a^n= nloga

After that we discussed that logarithms really just boil down to measurements and ratios.

Number 3 of activity 2.4 consisted of a girl named Emily who claimed that you can multiply any two power-scale numbers just by adding their corresponding logarithms.  She is indeed correct.  We can use log values you know to find ones that are unknown.  For example:

log6

log3 + log 2 = log 6

.301 + .477= .778

Then our class broke through and we learned a lot about multiplying power number.  The key is knowing that you can multiply the numbers by adding the logarithms.  Like we learned in earlier math subjects when you multiply numbers (powers), you add the exponents.

2^3+2^6 = 2^9

An example would be:

10 * 100

= 10^1 * 10^2

= 10^3

corresponds to…

log 1 + log 2 = 3

It would be easier to understand this to look at the example in the animation.  Click on the light blue link below to view it.  I know i didn’t explain this very well but I hope I helped a little bit.  I also sure that i messed up somewhere so please tell me so I can change it.  Reversearp can explain it a lot better so go ask him.  Please comment if you would like something cleared up.

2010-02-09_0956

P.S. the next scribe starts with a J and ends with an ocelyn.

The past two days in class, we have been working on Activity 2.4 in our packets. Activity 2.4 discusses the power scale versus the the exponent scale. By adding one on the exponent scale, we see that it is equivalent to multiplying by ten on the power scale. So, on the exponent scale, the difference [0,1] makes the difference on the power scale: [1, 10]. This is because 10^0 is 1, and 10^1 is 10. By realizing this, we can also prove that adding 2 on the exponent scale is equivalent to multiplying by 100 on the power scale. Therefore, we can say that 10^0 is 1, and 10^2 is 100. The difference between the two is a multiplication factor of 100.

As we moved on through Activity 2.4 we measured the distance from 1-2 on the power scale is 18 mm and the distance 0-1 on the exponent scale is roughly 59 mm. We divide 18 by roughly 59 to get .301. This proves that each multiplication by 2 on the power scale is adding another .301 on the exponent scale. log(2)= .301 which also means that 10^.301=2 because log(x)=y and 10^y=x. This lesson lead us to discover that log(a)^n = (n)log(a). We discovered this because a student pointed out that in order to find the log of 2^5 we need to move .301, 5 times down the exponent scale, because the exponent is 5. log(2^5) =5log(2). The distance from zero on the exponent scale is the logarithm of whatever number you’re looking for. We then began to work on a problem proving that you can multiply two power numbers by adding their logarithms. We didn’t get very far on this concept but so far we have seen: power 10*100 (and the logarithm of 10 is 1, 100, 2) is 1000=10^3. Adding their logarithms gives you 3, and 10^3 is 1000.

That just about sums up what we’ve done in class for the past two days. I can’t remember who the 3 people left on the list are so the next scribe is To Be Determined

Now that my computer is virus free, i can post my blog on logarithms.

So the past few days we’ve spent time using logarithms to find things like magnitudes and decibels. We’ve come to defining the logarithm as the exponent raised to a power of 10. For example:

log(100)=2 because 10^2=100

We’ve learned to find Magnitude of an earthquake as log(I/Io). (I/Io) is the relative intensity. That means you take the value for which you want to find the magnitude of over the smallest detectable earthquake which is given. Io= 2.00 x 10^11 in any case that deals with magnitude. Let’s say I= 2.518 x 10^18. We would have:

log(2.518 x 10^18/2.00 x 10^11). When you divide these values, you get an exponent of 7. Since the number you get isn’t a power of 10 raised to an integer, the log won’t be an integer. It is actually 7.1

We also got the problem:

How many times more powerful is the sound of a chainsaw (110dB) than the noise generated by a vaccum cleaner (sound intensity 10^-2)

We have to find the sound intensity of the chainsaw. In order to do this, we need the equation for dB

dB= 10*log(I/Io) Io= 10^-12. This is the smallest detectable sound by the human ear.

110=10*log(I/10^-12)

11=log(I/10^-12)

Since our answer is 11, we know that  the number inside the parenthesis will have a power of 11. We know the rule for dividing exponents is that you subtract them so x–12=11. x=-1

11=log(10^-1/10^-12)

The relative intensity for a chainsaw is 10^11

The relative intensity for a vacuum cleaner is (10^-2/10^-12)= 10^10

So we see that a chainsaw’s sound is 10 times more powerful than that of  a vacuum cleaner.

That’s it for logs- the next scribe izzz Steeenz Lyrixz AKA Eric

On Friday in class, Mr. B gave us a problem to start with.  It was as follows:

How many times more powerful is the sound of a chainsaw (110 db) than the noise generated by a vacuum cleaner (sound intensity 10^-1)?

Using the equation we know for finding the intensity of a sound, we can set up the equation:

110 = 10 * log (I/Io)

Since we know that the relative intensity (the sound of a whisper) is 10^-12, we can fill that in for Io, giving us this:

110 = 10* log (I/10^-12)

Next, we can divide by 10 on both sides to simplify:

11 = log(I/10^-12)

Here is the important part….Since the magnitude (11) is the number of the exponent of the log function, we know that (I/Io) needs to equal 10^11.

Next, we can evaluate for the vacuum cleaner with a similar process.

Since the intensity is is 10^-1, and the relative intensity is still 10^-12, you still get an answer of 10^11.

So, they are the same.

We also touched on the point that when making a number line for these values, the evaluated log function (such as magnitude in the earthquake example), the points will directly correspond to such points on a number line showing intensity. Hence, it is vital to make sure that the number line is scaled in a uniform manner.

Lastly, we found that logx=y is a logarithmic function, while 10^y = x is an exponential function.  This is why we have been studying logarithmic functions; they have a close relationship with exponential functions.

The next scribe will be…..named later, because I don’t have the list of people who have yet to go twice.  This is my third time, just so all of you know, so don’t pick me again.  EVER.

One love,

Nathan

Today in class we recieved a new packet. This packet introduces the concept of a logarithm function. The logarithm converts any positive number into its “power-of-ten exponent”. Today we did activity 2.3 in the packet through number 3. These problems used two new concepts:

magnitude: M=log(l/l0)

relative intensity: RI=l/l0

Tomarrow we will be going over these problems so everyone should have problems 1-3 completed.

The next scribe will be stefan.

Today in class we presented the claims for exercise 2.1. So I’m going to explain the problems that were presented and later I’ll try to clarify the concept of e in mathematics and how to use it.

Exercise 2.1 Practice Problems

3. A bank account paying 8% annual interest compounded quarterly actually pays 2% interest each quarter. The annual yield is slightly higher than 8% due to the compounding.

a) If $1500 is deposited when the account is opened, how much interest is earned during the first year?

If you took notes on Thursday, this problem is very similar to the first problem in activity2.2. So I first set up the equation 1500(1+.08/4)^n, because the interest is being compounded quarterly. Then to solve the problem, you must plug in the number 4 (because there are four quarters in a year), and get the account total being $1623.64824….

But that’s not the final answer for letter a, in which you have to subtract the total value you got from the initial $1500, to get the interest earned. And the final answer is 123.64824….

b) What is the annual yield?

For this they are looking for a percentage, and to find that percent you take your interest earned in letter a and divide it by the initial amount, $1500. Like so,

123.64824….(interest)

1500(initial total)

And the final answer for b would be 8.24….%

c) If the money is invested for a 5-year period, what will the Balance be at the end of that interval?

This question is similar to letter a, and all you have to do is plug in twenty, (number of quarters in five years) for n, into the equation 1500(1+.08/4)^n.

The final answer would be $2228.92….

4. As previously noted, if you deposit $1 in a bank account paying interest at an annual rate of 100% compounded continuously, you would end up with e dollars after one year.

The Constant e

Well, to first understand this question, I may need to clarify the concept of e.

To better understand e you can compare it to pi, something we already know. It’s a constant number that many people try to memorize the consecutive digits, like pi.

The number e frequently occurs in mathematics and is an irrational constant (like ?). Its value is

e = 2.71828182845904523536028747135266249775724709369995…

The number e is used as a limit to how much some can be or how little it can be. It can also be represented in a graph with an asymptote because a value has only potential to reach that e value but can’t exceed it. Like in the following table, the values reach a point, but doesn’t exceed past that point.

Here’s also a website to further understand the number e. http://en.wikipedia.org/wiki/E_%28mathematical_constant%29

Go to the compound interest problem section,  it explains similar problems that we did in class.

a) With continuous compounding, how much would be in the bank after two years?

This would be represented with the expression e^2, because it’s continuously compounding which is e and it does so for 2 years which is the exponent. So the final answer is 7.389….

b) With continuous compounding, how much would be in the bank after five years? After t years?

Just as before, this would be represented with the expressions e^5 and e^t. The final answer for e^5 is 148.41….

c) Use your calculator to find 100*e^(.08). How does that answer compare to the work done in item 1 of activity 2.2?

The answer on my calculator was 108.32…. which was the same as the limit in the table, which can be referenced above.

d) Review your answer to item 3 of activity 2.2 and this exercise. Then generalize that work to write an expression for the balance after A dollars at 100r% compounded continuously for t years. Use numbers to check your expression for a specific case.

The generalized expression I wrote for this was b=A*te^%

Hopefully that helped clear up any questions you had about e and problems involving it, and if you still have some questions you can look on the website previously mentioned or you can read the section Base e in the packet on page 87. But if your more of an auditory or visual learner here’s a video on youtube that can maybe help. But as a warning: this video is boring, but informational. So if you’re not understanding the subject I recommend it.

http://www.youtube.com/watch?v=dzMvqJMLy9c

The next scribe will be Heather.

Today in class Mr. B gave us a problem involving exponential functions to work on. This problem stated:

Function f has values f(5)=12 and f(10)=18. Find f(20) and f(x).

First thing we did was create a table. In the X column was 5, 10, and 20. In the f(x) column was 12, 18, and a blank, since we needed to find f(20). Then we used our Add-Multiply method to find b in our a*b^x equation. For this, Mr. B showed us a time saving method we can use for this step. First, we create two equations using the values we know: 12=a*b^5 and 18=a*b^10. Next, we begin substitution to get a=12/b65 and a=18/b^10. Finishing off the substitution method, we are left with one equation: 12/b^5=18/b^10. When we simplify, we get b^5=1.5. Then we find the 5th root of 1.5, and tada! We have our b value for our exponential function equation: 1.08… To find a, we can use the values we already know to set up an equation. This equation is 12=a*1.08…^5. When we solve this out, we find that a=8. Now we have all we need to form our equation. f(x)=8*1.08…^5.

After we finished this, we were given the rest of class to work on claim problems for tomorrow. Mr. B said that he will choose the claim problems from 1-9 in section 2.1 of the packet, so be ready to present tomorrow. The next scribe will be Amie! =D

During class on Friday, we spend the hour solving the problems in exercise 2.1 of the exponential functions packet. We will have some class time on monday to continue and claims will be on Tuesday.

Some tips for solving exponential function problems:

-know how the equation y=ab^x works and understand the restrictions: a cannot equal 0, b cannot equal 1, and b must be greater than 0

-understand growth rate (next/current) and relative rate (the change in percentage of the amount present)

-know that Euler’s number is the limiting value of exponential functions and know how to use it in problem solving

-create charts like the ones in the packet if you get stuck

-also, table 2.4 may be helpful in understanding the nature of change in certain situations in exercise 2.1

Good luck! The next scribe will be Emily M.

Today in precalc, we spent most of the time working on problems out of a packet that deals with exponential functions. Because we learned the basics about exponential functions yesterday, that means that today we applied these concepts.

The problem we had to solve today was to determine if there was a limit to the amount of money that would be earned in a savings account if we continually added a compound interest of 100%. The expression given to us (a form of an exponential function) was (1+1/n)^n.

First we had to make a table.

If you look at the last few values, you might notice that they have seemed to reach a peak at about 2.72. This value is actually an irrational number (like pi) that is called Euler’s number. The symbol for this value is e. The exact value (or as close as I can get) is 2.718281828… The formal definition of e is (1+1/n)^n.

So when applying this concept to banking, this means that if you want to make the most amount of money by having the interest compiled more frequently, there is actually a limit at which the final amount will not increase at a significant rate.

Basically, today was a day where we worked on a few problems and continued our study into exponential functions. The next scribe will be…Shannon. Be happy. You have the entire weekend.

Today in Pre-Calc, we spent most of the class working on Activity 2.2 in our packets, problems 1-3. These problems dealt with another type of exponential function:

f(x)= (1+1/x)^x

The first problem dealt with a real life application problem dealing with compound interest. We were told to complete a table for the function:

f(x)=100(1+.08/x)^x

We needed to figure out the values for your balance annually, semiannually, quarterly, monthly, daily, and hourly. Here are the values for each of the time periods:

Annual: $108.00; Semiannual: $108.16; Quarterly: $108.24; Monthly: $108.30; Daily: $108.33; Hourly: $108.33

The basic pattern in this function would be it increases at an exponential rate but after a certain time period, it will begin to level off. The number that this function begins to level off at would be $108.33. These types of functions are used quite frequently by almost every banker.

To figure out how to get there, we built up the function. We started off by creating a table and a graph for the function:

f(x)=1/x

That graph made an L-shape in the first quadrant. There was a horizontal asymptote at the y=.01 line. This is because the f(x) value will never get to .01, it will just approach it. Then, we derived another function, which was:

f(x)=1+1\x

This graph also produced an L-shaped curve in the first quadrant. However, the horizontal asymptote instead of being at .1, was at 1.01. This was because it will approach 1.01, but never quite reach that value. This leads us to the function I mentioned earlier:

f(x)=(1+1/x)^x

Unlike the first two graphs, this one produced a half-upside down bowl shaped curve. Sorry for the explanation of the shape. Anyways, this graph is shaped like this because your number will start low and gradually climb higher due to the exponent. The “limit” of this function is approximately 2.718….. This is actually a special number many bankers use, and its symbol is e.

On a side note, we also spent a good part of the class period talking about the new iPad and doing some random background checks. Fun times. Anyways, tomorrow’s scribe is ……………….. Emily M.