Precal Blog

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.

Today in honors Pre-Calculus hour 3, we discussed exponential functions. First we stated the basic equation for exponential functions can be written as f(x)= a*b^x. We further determined that exponential functions have certain “rules” they must follow, they are b>0, b can’t equal 1, and a can’t equal 0. For the first rule, if we use a number less than 0, like -2, our answers for the f(x) value would alternate from positive to negative. For the second rule, if b equals 1, we would simply get back the a value leaving us with a straight line. For the final rule, if a is equal to 0, our answer is 0 giving us nothing. With these rules in mind we looked at four different exponential graphs. I don’t have images so I’m sorry you must bear with my descriptions. The first graph was a>0, b>1. The graph was a positive exponential graph starting above the x-axis getting larger. The second graph was a<0, b>1. The graph was negative starting below the x-axis and sloping down into the fourth quadrant. The third graph was a>0, 0

Next we went over the 2.1 packet problem 3. We decided the domain for an exponential graph could be all real numbers, but the range had some stipulations. If a is negative, y<0; if a is positive, y>0. After testing the f(-x)=f(x) and f(-x)=-f(x) equations for even and odd symmetry, we decided that exponential graphs fit neither so they have no symmetry. Next we deduced that exponential graphs can increase or decrease everywhere, across the entire domain. Finally we looked at the concavity of exponential graphs. graphs can either can be concave up (“u-shaped” so they hold water) or concave down (upside down “u” so they shed water). In the case of exponential graphs if a>0, the graph is concave up; if a<0, the graph is concave down.

The last concept we discussed was “end behavior”. I’m not entirely clear on this concept (Mr. B said he struggled with it in high school so I don’t feel that bad) but basically it evaluates towards what “number” the graph is going. In the case of graph one from above the x-values are “going towards” positive infinity; and the f(x) values are also “going towards” positive infinity. When analyzing end behavior though you must make two sets of…”it”. One for the positive side and one going towards the starting point. So we have our positive side, now we look at the side going towards our starting point. in this case of graph one we have our x-values going towards negative infinity, and f(x) values going towards zero. Sorry if that was confusing, math terminology is beyond me so that’s my best attempt. We finished class by getting page 89 of the packet and to work on those problems. The next scribe will be Jake.

Today in class we talked more about Exponential functions, and went through problem 3 from the worksheet we got yesterday.
You can graph Exponential functions in four different ways:

F(x) = ab^2  <—— general equation

The b-value represent the stretch
b > 0
b -> not equal to 1
a -> not equal to 0

Domain/Range
D: all real numbers
R: if a > 0 then y > o
     if a < 0 then y < 0

Symmerty –> None
* no even symetry
* no odd symerty

Increasing/Decreasing
* always increasing or always decreasing
* always one-to-one function

Concavity
If a is positive = concavity up
If a is negative = concavity down

Inflection points:
Its the points where it change from concavity up to concavity down. And Concavity down to concavity up.

end Behavior
( in the picture on top)

The next scribe is Riley