Monday, November 12, 2007
Saturday, November 10, 2007
Logarithms: Change of Base Law
Hello fellow students and guests, this is Lina bringing you the scribe post for Friday’s class. In the morning, half the class was missing, but as they say, life goes on and we started the class by solving logarithms.
It is a formula that allows you to rewrite a logarithm in terms of logs written with another base. *As defined by my friend, mathwords.com In other words, it allows you to solve an equation that has different bases as shown later on.
It finds for you the exact exponent. When using the change of base law and no base is indicated, it is always “base 10” which is the common logarithm or common law. Since your calculator only allows you to calculate base 10 logs, converting your logarithm into the fraction as seen above automatically changes it to a base 10 log making it easier to calculate. Therefore, it lets you solve any logarithm on your calculator, but keep in mind that you will not always have a calculator for these types of questions. I’ll elaborate on that later.
The reason given to us for why we calculate in Base 10 is simply because we have ten fingers and it is easier to count. If we were aliens with six fingers, then we may just use “base 6”, but let’s not get confused here.
So our first task during the morning class was to solve the following using logarithms.
Our first question:
While keeping in mind from previous lessons that a logarithm is indeed an exponent.
As known in this:
We proceeded to solve the question.
-First, we converted each side in terms of logarithms.
-Also remembering the power law, the exponent “x” is placed in front of the log3
-We then divided both sides by “log3” to isolate the “x” on one side.
-Then we inputted the “log12/log3” on our calculators to get our decimal approximation for “x”.
Our second question was very much the same.
*Even though the brackets aren’t given around the exponent in the original question, you should always put them in for safety measures as not to get confused later on in solving the question.
The answer written in blue is given by a student. Following the same idea as in the previous question and we ended up with the decimal approximation. However, Mr. K showed us a different way of solving the question from the second line in blue where (x+ 1) log3 = log 17.
-From that line, we factored out the (x + 1) out giving us xlog3 + log3 = log17
- We bring the log3 to the right side of the equation
- We divide both sides by log3 isolating the x.
-Badda bing, Badda boom we have our answer in the exact form of: x = (log17 + log3) / (log3)
Usually you would stick to writing the exact answer when you are solving this sort of problem during the non-calculator part of the test or exam. You can write a decimal approximation if you are allowed a calculator, although it’s not necessary unless it specifically states in the question that you have to write out the decimal approximation.
Somewhere in the middle of the lesson Mr. K started talking spontaneously about Base Numbers and how you would make them. For example as seen in the above in Base 2, the symbols used to make up numbers in Base 2 are “0” and “1” only. So the numbers one through four are written using those said symbols only. It all has to do with place value.
I’m not quite sure I can explain this concept well, but here’s a website that you may find interesting that elaborates on this idea. http://www.purplemath.com/modules/numbbase.htm
It’s not that important to know as of now, but you may want to check it out.
By the afternoon class, the missing half of the class joined us once again and we were greeted with yet another pop quiz. My fatal mistake was forgetting for that brief moment that a logarithm is an exponent!
After correcting the quiz together, we proceeded to solve more questions after Mr. K caught the missing half of the class up on what occurred during the morning’s lesson. They caught on quite quickly I must say.
Here’s a sample of one of the questions we worked on. It follows up on the things we learned that morning.
They were solved differently, but are nevertheless the same answer in the end. However, the answer on the left is a little more expanded than the one on the right. You can write it either way as they are both correct.
So somewhere in the middle of both morning and afternoon classes, Mr. K decided to share a joke with us.
We all know the story of Noah’s
So the second part of the joke. Noah walks to the newly formed forest and after hearing some *crash crash* *smash smash* *bang bang* *hammer hammer*, Noah comes back and once again says to the snakes, “Go forth and multiply!”, but again the snakes look puzzled and say they can’t. Noah then says, “Of course you can because I’ve made you LOG TABLES.”
Well that’s all from me for today. So last, but not least I will name our next scribe. JESSICCA!
Friday, November 9, 2007
Thursday, November 8, 2007
Logarithms
Hey guys, it's Kimberley and I am today's scribe.
Well basically all we did today was expanding logarithms, simplifying logarithms and what Mr. K called a "classic question" (on an exam that is).
To begin, EXPANDING LOGARITHMS:
Here is an example that we spent a very long time "debating" over for the first part of the class.
We were asked to expand as much as possible. In order to do that, we must apply the logarithm laws.
Let's begin with analyzing only the top portion of the logarithm. As you can see above, everything is in terms of log base a. Now the product law states that when powers are multiplied, the exponents are added. Above we have the power of "A" being multiplied by the power of "the root of C", by the product law, we are able to expand this by adding the exponents (*remember, LOGARITHMS are EXPONENTS). So we now have:
Now let's consider the bottom portion of the equation. The quotient law states that when dividing powers, we subtract exponents. The original question is dividing by the power of "B^2", so this means that according the quotient law, we are able to subtract from log base of b. Now we have:
Lastly we need to apply the power law. The power law states that when something is to an exponent, to an exponent, the exponents are to be multiplied by one another. Following this rule, we come up with the expression:
Now here is where all the "debating" began. Some students said that 2 multiplied by log base A of B can be expanded even further, which is correct, but that would also mean that 1/2 log base A of B can also be expanded further. If we were to continue, we could also expand it again even further, this means that it can be expanded infinitly. So the question is, when is it enough? The answer is, for the purposes of exams and such, the answer is complete once we have applied all the possible logarithmic laws. Because all the laws have been applied to the question above, it is considered to be complete.
Now we will look at SIMPLIFYING EXPRESSIONS:
This did not require as much time to complete as the expanding questions because everyone seemed to understand it pretty well, so i will not go into great depth in explaining it.
Here is an example. We were asked to simplify this into a single logarithm.
Basically all you need to do is apply the logarithm laws in reverse. Where exponents are being subtracted, powers are being divided. Where exponents are being added, powers are being multiplied. When exponents are being multiplied by one another, the power is to an exponent. As you can see above, everything is in terms of log base 2 of A, so we can combine everything quite easily. The power of "A" is multiplied by the power of "C", because of the power law. All of "A" multiplied by "C" is then divided by "B", due to the quotient law. We also have log base 2 of B being multiplied by 2, and log base 2 of C being multiplied by 1/2, so we cane use the power law to simplify this. Once we have applied all the logarithm laws, we come up with:
Lastly, we will look at the "CLASSIC QUESTION"
This is a typical question that you may see appear on a exam.
What you need to do is manipulate the logarithm so that you can use the given values. For example, 15 can also be written as 5 x 3. You will then have:
Now we can apply the product law to rewrite the logarithm once again. When multiplying powers, we add exponents. So instead of multiplying 3 and 5, we can add log base a of 3 and log base a of 5:
Basically all we have left to do is input the values of log base a of 3 and log base a of 5 that were given earlier in the question. Once we do that, we will have:
Well that is all we did today. Other examples can be found on the slides that were posted by Mr. K. Homework for tonight I believe is Exercise 23.
Oh by the way, the next scribe is MILLER (:
Well basically all we did today was expanding logarithms, simplifying logarithms and what Mr. K called a "classic question" (on an exam that is).
To begin, EXPANDING LOGARITHMS:
Here is an example that we spent a very long time "debating" over for the first part of the class.
We were asked to expand as much as possible. In order to do that, we must apply the logarithm laws.Let's begin with analyzing only the top portion of the logarithm. As you can see above, everything is in terms of log base a. Now the product law states that when powers are multiplied, the exponents are added. Above we have the power of "A" being multiplied by the power of "the root of C", by the product law, we are able to expand this by adding the exponents (*remember, LOGARITHMS are EXPONENTS). So we now have:
Now let's consider the bottom portion of the equation. The quotient law states that when dividing powers, we subtract exponents. The original question is dividing by the power of "B^2", so this means that according the quotient law, we are able to subtract from log base of b. Now we have:
Lastly we need to apply the power law. The power law states that when something is to an exponent, to an exponent, the exponents are to be multiplied by one another. Following this rule, we come up with the expression:
Now here is where all the "debating" began. Some students said that 2 multiplied by log base A of B can be expanded even further, which is correct, but that would also mean that 1/2 log base A of B can also be expanded further. If we were to continue, we could also expand it again even further, this means that it can be expanded infinitly. So the question is, when is it enough? The answer is, for the purposes of exams and such, the answer is complete once we have applied all the possible logarithmic laws. Because all the laws have been applied to the question above, it is considered to be complete.Now we will look at SIMPLIFYING EXPRESSIONS:
This did not require as much time to complete as the expanding questions because everyone seemed to understand it pretty well, so i will not go into great depth in explaining it.
Here is an example. We were asked to simplify this into a single logarithm.
Basically all you need to do is apply the logarithm laws in reverse. Where exponents are being subtracted, powers are being divided. Where exponents are being added, powers are being multiplied. When exponents are being multiplied by one another, the power is to an exponent. As you can see above, everything is in terms of log base 2 of A, so we can combine everything quite easily. The power of "A" is multiplied by the power of "C", because of the power law. All of "A" multiplied by "C" is then divided by "B", due to the quotient law. We also have log base 2 of B being multiplied by 2, and log base 2 of C being multiplied by 1/2, so we cane use the power law to simplify this. Once we have applied all the logarithm laws, we come up with:
Lastly, we will look at the "CLASSIC QUESTION"This is a typical question that you may see appear on a exam.
What you need to do is manipulate the logarithm so that you can use the given values. For example, 15 can also be written as 5 x 3. You will then have:
Now we can apply the product law to rewrite the logarithm once again. When multiplying powers, we add exponents. So instead of multiplying 3 and 5, we can add log base a of 3 and log base a of 5:
Basically all we have left to do is input the values of log base a of 3 and log base a of 5 that were given earlier in the question. Once we do that, we will have:
Well that is all we did today. Other examples can be found on the slides that were posted by Mr. K. Homework for tonight I believe is Exercise 23.
Oh by the way, the next scribe is MILLER (:
Wednesday, November 7, 2007
Logarithm Laws
Hey its Anthony. Im the scribe...again. Ok today we started the morning class off with getting back all our old work and tests. My oh my was that a heartbreak, but then again it really was a wake up call to reality. So here I am, trying to do better. Back to topic we did our usual warm up questions and we got introduced to a new idea. Actually three new Laws. The Product Law, The Quotient Law, and the Power law.
Now here are some steps to use each of the following laws. We are going to start with the Product law since it the was the first that was introduced to us.
The product law. *When we use these laws we can look back to the log formula for reference.
The formula is as follows:
b = base
c = power
a = exponent
Example question *taken from todays slides*.
1)So we start by expanding. Then we get the first line.
2)Secondly As we know 2 is the base and 8 and 16 are the power. Now we ask our selves. "2 to the power of what equals 8 and 2 to the power of what equals 16". So we get 2 to the exponent 3 equals 8 and 2 to the exponent 4 equals 16. Right? check your calculator!
3)Now to refer back to our Product law, when adding 2 logs together we multiply the exponents. Now back to the question, we have now established 3 and 4 are the exponents and 2 is the base. Now the hard part 3 + 4 = ?
The second Law introduced to us was the Quotient Law.
Example question *taken from todays slides*
1)Ok again lets start with expanding to get the first line. Ok now we refer to the quotient Law. "When taking the difference of 2 logs we divide Log M by Log N."By doing this we M = 32 and N = 128. Since the Law is M/N. Now exspanding we get line 1.
2)Next we figure out the exponents and the base. To do that we refer back to the Log formula. So now we are at the base is once again 2 and we ask ourselves the same question. "2 to the exponent of what equals 32 and 2 to the exponent of what equals 128." I believe 2 to the exponent of 5 equals 32 and 2 to the power of 7 equals 128. Check your calculators!
3)Now we have the exponents 5 and 7. Plug it into the Equation and Tada! The answer is negative 2.
Lastly The Power Law.
I found this law the most difficult to grasp. As Mr. K said. " This is what students get upset about, its so simple that its difficult"
Example question *taken from our slides today*.
1)Ok again lets start by Determining the base and exponents. So by looking at the question and refering to the Log formula the base is 2 again... and the power is 8. Lets asks ourselves "2 to the exponent what equals 8?" I believe it is 3. Check your calculators!
2)So now we have 2 to the exponent 3 IN BRACKETS to the exponent 5. So what is the rule when you have powers to the exponent of whatever? You multiply exponents. Ok back to the question. Remembering the rule we get 3 multiplied by 5. So we get an answer of 15. Sweet.
THE NEXT SCRIBE IS ............ KIM - POSSIBLE !!!
Now here are some steps to use each of the following laws. We are going to start with the Product law since it the was the first that was introduced to us.The product law. *When we use these laws we can look back to the log formula for reference.
The formula is as follows:
b = basec = power
a = exponent
Example question *taken from todays slides*.
1)So we start by expanding. Then we get the first line.
2)Secondly As we know 2 is the base and 8 and 16 are the power. Now we ask our selves. "2 to the power of what equals 8 and 2 to the power of what equals 16". So we get 2 to the exponent 3 equals 8 and 2 to the exponent 4 equals 16. Right? check your calculator!
3)Now to refer back to our Product law, when adding 2 logs together we multiply the exponents. Now back to the question, we have now established 3 and 4 are the exponents and 2 is the base. Now the hard part 3 + 4 = ?
The second Law introduced to us was the Quotient Law.
Example question *taken from todays slides*
1)Ok again lets start with expanding to get the first line. Ok now we refer to the quotient Law. "When taking the difference of 2 logs we divide Log M by Log N."By doing this we M = 32 and N = 128. Since the Law is M/N. Now exspanding we get line 1.
2)Next we figure out the exponents and the base. To do that we refer back to the Log formula. So now we are at the base is once again 2 and we ask ourselves the same question. "2 to the exponent of what equals 32 and 2 to the exponent of what equals 128." I believe 2 to the exponent of 5 equals 32 and 2 to the power of 7 equals 128. Check your calculators!
3)Now we have the exponents 5 and 7. Plug it into the Equation and Tada! The answer is negative 2.
Lastly The Power Law.
I found this law the most difficult to grasp. As Mr. K said. " This is what students get upset about, its so simple that its difficult"
Example question *taken from our slides today*.
1)Ok again lets start by Determining the base and exponents. So by looking at the question and refering to the Log formula the base is 2 again... and the power is 8. Lets asks ourselves "2 to the exponent what equals 8?" I believe it is 3. Check your calculators! 2)So now we have 2 to the exponent 3 IN BRACKETS to the exponent 5. So what is the rule when you have powers to the exponent of whatever? You multiply exponents. Ok back to the question. Remembering the rule we get 3 multiplied by 5. So we get an answer of 15. Sweet.
THE NEXT SCRIBE IS ............ KIM - POSSIBLE !!!
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