September 23, 2006

logarithms assignment, group 3

Group 3: Suzanne, Christian, Mark


We found out that this equation is true, here is our solution...

One of the laws of logarithms is: log(M/N)=logM-logN, also known as the Quotient law .

Using the Quotient law, ln(A/B)=lnA+ln(1/B), the blue part can be simplified to ln(A/B)=lnA+ln1-lnB.


If you solve for ln1, it equals 0. So our equation, ln(A/B)=lnA+ln1-lnB
simplifies to

ln(A/B)=lnA-lnB The equation holds true, if you go back to our Quotient law (log(M/N)=logM-logN).

ln(A/B)=lnA-lnB can be simplifed to ln(A/B)=ln(A/B)


ln(A/B)=lnA-lnB can be simplified to lnA-lnB=LnA-ln B

Here is a possible mistake a person might make when writing this equation as:


Instead of using the quotient law correctly, someone may subtract ln(1/B). IF they subtract like this, they have made their first mistake.>>>ln(A/B)=lnA-(ln1-lnB)


The right side of the equation results in an addition, WHICH IS DIFFERENT FROM THE QUOTIENT LAW. ln(A/B)=lnA+lnB

The above equation would simplify to this:

ln(A/B)=ln(AB) SEE THE DIFFERENCE?on the left side the values are being divided. On the right side, the values are multiplied

The reason someone might do this is by mistakenly applying the logarithmic law log(M/N)=logM-logN to this question twice.


jann said...

I agree with your answer group 1. Since the ln(1/B) is the same as ln(1)-ln(B)and the ln(1)=0, the equation is true. The law ln(A/B)=ln(A)-ln(B) implies. Nice job.

lindsay said...

i agree with your answer. however, i'm having a hard time understanding the solution maybe it's just me, i just don't really follow. XD. but i do think that it's true. =)

be happy.

linger said...

I agree with Lindsay, that the explanation is not easy to follow. Maybe plugging values in would help (it'd help me anyway) Ohh and can you please make the font a little bigger, it's kind of hard to read. I don't mean to be picky lol, I'm just saying. But other than that, good job =)

lindsay said...

i also made that mistake when solving an equation like that one. i actually thought it was true at first! but yes, now i know that i was wrong and i will try to never make that mistake again.

anh said...

I also found it hard to follow your solution, but I read it a couple of times so I understand what you did. I think that taking the ln(1/B) and using one of the laws to change it to ln1-lnB was a good way to prove that the equation was true, so I agree with your answer.

M-A-R-K said...

i was trying to post it directly from writely but i was having problems. so i had to copy it and paste to the blog. It took all the spacing between each problem. if you remember the quotient law our solution is easy to understand. i'll try to edit it and explain it a bit more. If i can figure out how to post from writely, i will work with my group to make a more understandble solution =)
sorry for any inconvenice,
group 3 staff lol =p

crystal said...

I was also a bit confused with your explanation. But after I wrote it out myself and read it over more, it made perfect sense. Your groups answer is perfect for an exam or test. The steps and colors were very helpful. Great use of the Logarithmic Laws. Good job!

ashlynn said...

I agree with others about having to read over the explanation again. I had to think it out myself for a bit, but your explanation totally helped me. I think you guys did a great job. Plus your use of colours also helped.

char__lene said...

i agree with what other people said, i kind of had to thing about the solution but after really reading it i do understand why you guys how your answer. i think the colors make it easier. good job

Manny said...

Your answer is right, it is true. But the explanation is a bit choppy. Use the product law, it's more easier to understand.

When adding exponents of the same base, you multiply the powers. Therefore...

ln(A/B = ln(A/B)

Your other explanation works, but it just takes longer =). Good job.

Suzanne said...

That's an excellent point manny.. I hadn't thought of that. It is simpler than what we did.

christian said...

YEA! AHAHA the solution to this question is much simpler than we thought. To make what this group came up with simpler, what throws people off initially is the awkward placing of ln(1). I guess what our break through is, is that ln(1) = 0 (what's the surprise there), hence it's just nothing. Don't mind it. That's all folks!

M-A-R-K said...

lol thanx Manny =D for seeing the product law. At least now there are almost 2 solutions for each question.

i just remembered something... with logs and exponents there are many possibilities =), what Mr. K said last year.

katrin said...

Although the explanation was a bit hard to understand, I still had an idea about why it was True and not false. I agree with your answer and I agreee that the product law is much easier to prove why the equation, ln(A/B)=lnA+ln(1/B, is TRUE. Overall, nice usage of color coordinationg the fonts, it really helped!

danny said...

I agree with your answer, however there are a couple of things I'd like to point out. First of all, although your solution does seem right I find myself having a hard time understanding your explaination for it, don't get me wrong, it's organized just not explained with enough detail. Second, the solution was posted very small maybe possibly to save space? It would've been easier to see if you were to make the text bigger. Even though, your answer was totally right and great job on that!

charlotte said...

GOdd job gys we got the same answer! I like the fact that u gys expanded the right side of the equation. It help me get the picture. I always tend to solve the equaton by pulgging in values for the variables, but know i know better!
Why does so many people had a hard time understanding ur solution, well for me i got it. THe colors truly helped!
Many thanks for pointing out the product law, it works well and your brilliant. I want to give u a path on the back. Well done

Amit said...

I don't know if this thread is active. Perhaps a simpler way is to treat this as Ln(A/B) = Ln[(A)(B^-1)] which reduces to Ln(A) + Ln(B)^-1 which is then equal to Ln(A) + (-1)Ln(B) which is then equal to Ln(A) - Ln(B)