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# May 2014 Math SL Study Help - Mixed Questions

Hello there! So I've been struggling with studying for this class, and it's been driving me nuts! You see, I took this class last year, but I had my schedule a little too packed and I really neglected the homework and such. Needless to say, I got a less than satisfactory grade on the final exam. So this year, I'm retaking it. Every day, or as many consecutive days and I can manage up to the date of the test, I'm going to post a math question with a worked example, or as worked as I can get it, asking for assistance. This way, if I get the answer correct, it can help anyone else out there struggling, and if I don't, someone else can hopefully provide insight?

I also encourage anyone who is working on math questions in a similar manner to submit their own that they haven't been able to figure out. Thanks in advance for all the help!

Chapter 25 - Miscellaneous Questions

Part A - Non-Calculator Questions (page 652)

1. A geometric sequence has S1 = 2 and S2 = 8. Find:

a the common ratio r
b the twentieth term u20.
Solution: First, common ratio r can be found by knowing the equation for the nth term of a geometric sequence, which is un=u1rn-1
We can now substitute in what we know above, if the sum of the first term is 2, then u1 must also be 2. We can find the sum of the first two terms now, by subtracting u1 from S2, 8-2=6, so u2 is 6. Now we substitute this into the formula for the nth term given above. 6=2r, therefore r=3
Now we know what r is, and we can find any term we want using the same formula. u20=2(3)19
This is a no calculator question, so the answer does not need to be simplified any more than it is above.
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2 Show that the sum of the first forty terms of the series ln2 + ln4 + ln8 + ln16+... is 820ln2.

Solution: For this question, it is not stated whether this is a geometric or arithmetic sequence, so we must figure this out. In an arithmetic sequence, the numbers would increase at the same interval, and in geometric, they would increase by the same ratio. In this one, they are increasing by the same ratio, and that ratio is 2. (Ratio can be found by dividing a term by the term before it, no matter what two consecutive terms you use you'll get the same result.)

Now, we use the sum of a geometric sequence formula, as show below, and substitute in what we know.

So here's where I start having an issue. As you can see in the file attached, I know what process to use, but I'm not very good at simplifying past that point. What do I do next? Reminder: This is a no calculator question.

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So no one ever responded to my question yesterday, so I'm going to try and solve it again today, I'm using a reference on solving logarithmic equations from purplemath, I have the link here.

So yesterday, I had simplified my answer down to 820ln2 = ln2(240-1)

First off, I need to make the ln terms the same.

I can change the term in parenthesis on the right side two ln terms instead, ln2+(40ln2/1)

Now my equation looks something like this: 820ln2 = ln2 + (40ln2) Right?

I think I'm getting a little stuck again, is this next step correct?

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I think you got the first question right.
For the second question, the problem you have is that:
$\ln{4}=\ln{2^2}=2\ln{2}$
Similarly,
$\ln{8}=\ln{2^3}=3\ln{2}$
So this is actually an arithematic sequence, not a geometric one.
The sum of the first forty terms is therefore $\ln{2}+2\ln{2}+3\ln{2}+\dots+40\ln{2}$
Which is
${(1+40)\times 40 \over 2} \times \ln{2}=820\ln{2}.$
By the way, not everyone is using the same book as you are, so if you could post which book yours is, that would be helpful.

Edited by ssy
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Ohh thank you so much! And I will do that. Here is what the front page of my book looks like, it is also really easy to find free PDF files of this book if anyone is interested in using it.

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So here's my next question. I used a screen shot because I'm not sure how to make some of the symbols and such shown. ^^;

Solution: The o symbol in these problems means 'of' so the second term is substituted for the x in the first term. A is therefore shown below:

(f o g)(x)=beln(bx)

This is not fully simplified, because the ln, as a power of e, will cancel it out. For instance: eln1 = 1 So for this problem, eln(bx) is just bx, so (f o g)(x) = b(bx) or b2x.

For (g o f)(x) we substitute the entire term of f into g.

ln[b(bex)]

So where do I go from here? I have a hard time simplifying these types of equations.

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I've done up the solutions to the questions for you - you can find them here. http://i.imgur.com/Hv0BP1t.jpg

For b, use log laws to simplify.

For c, equate the two and then solve for x.

Edited by Rahul
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Hi guys. I'm reviewing for the Math SL exam by going through some old papers and I encountered a problem in M08/5/MATME/SP2/ENG/TZ1/XX+ that I wasn't quite sure how to do.

The problem: http://someimage.com/yAwMeoa

I do know the process: evaluate an integral to find the area but I'm confused as to how to find the intersection points of ln and cos, as I don't know how to graph both types of functions at the same time in my Casio calculator. If I try, it comes out funky.

Edited by nawalnawal8

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I'm afraid your calculator is the only way to do this. Try window settings x:[1,10,1] and y:[0,7,1], and put y1=f(x) and y2=g(x). Let me know if you can calculate the interception points from that.

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I can't put both in at the same time because the ln function will not show up if I set the view window to trig to allow the cosine function to show up.

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You don't have to set the view window to trig. Just use the window settings I put in my last post: x:[1,10,1] and y:[0,7,1], y1=f(x), and y2=g(x).

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Awesome, it worked. Thank you!!

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