Precal Blog

Today, as in Friday, we learned more about logarithms  and the properties used with them. The three we learned were

log(a^n)=nlog(a)                      log(2^6)=6log(2)

log(a)+log(b)=log(a X b) log(2)+log(3)=log(6)

log(a)-log(b)=log(a/b)            log(10)-log(5)=log(2)

We derived these using different examples like the ones shown above.

Then we talked about using theses properties in equasions like

log(x)-log2(x-3)=3  —>  log((x-1)(x-3)) —> 2^3=(x-1)(x-3) —> x=5

Lastly, we reviewed teh logs we already know and practiced using them to fill in a number line.

log(4) = log(2) + log(2) = .602

log(5)= log(10/2) = log(10) – log(2) = .699

log(6) = log(3*2) = log(3) + log(2) = .778

etc….

Thanks(:

Today in class we reviewed what we learned yesterday and added some new log equations that can help find answers easier. Also, we looked at figuring out logarithms for 4 – 10 using the equations.

The logarithm equations (?) that we now have are:

log(a^n) = n log(a)       ex. log(2^5) = 5 log(2)

log(a) + log(b) = log (a*b)     ex. log(10) + log(2) = log(20)

log(a) – log(b) = log (a/b)      ex. log(30) – log(5) = log(6)

You can use the powers 10 scale to figure out the following problems or solve them using the equations above as I have them shown below:

log(4) = log(2) + log(2) = .602

log(5)= log(10/2) = log(10) – log(2) = .699

log(6) = log(3*2) = log(3) + log(2) = .778

log(8) = log(4*2) = log(4) + log(2) = .903

log(9)= log(3*3) = log(3)  + log(3) = .954

You can also use the equations to figure out more complicated logarithms:

log(8/3) = log(8) – log(3) = log(4) + log(2) – log(3) = .426

Hope this helped! The next scribe will be Steph Q.

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