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View Full Version : "Linearising Using LOGS"- This question is too hard! Help!

Farvahar021
06-18-2011, 03:30 PM
Well I've been give bunch of assignments to do and I have finished all of them but this one :shock: ... I have no idea at all how to do this, I know the basics of Logs (up to AS level in UK) but this question is just too hard for me, I hardly even understand the question :S please help

I've scanned the question and uploaded it on image shack website
http://img23.imageshack.us/img23/863/mathsqm1.jpg

Thanks and I hope someone here can help me out.

galactus
06-18-2011, 04:19 PM
Do you have Excel?. It will do this lickety-split.

Use Excel to perform a Exponent Regression with the given data.

I got I=.02e^{\frac{-t}{\frac{1}{5}}}=.02e^{-5t}

Parts b and c are easily answered now.

Do you know how to make a chart like this in Excel?. If so, adding a trendline is easy.

Farvahar021
06-18-2011, 05:40 PM
Umm no im not very good with excel but i think i can manage to make a chart using my number but how does that help me out i dont understand.

after I saw "Exponent Regression" in your post i didnt even know what it was and I just did some research on it and I realised that my teacher hasnt even been over this subject.. I mean Logs yes but all this is new to me. Could you tell me how you got 0.02 = I and 1/5=T?

http://people.hofstra.edu/Stefan_Waner/ ... ssion.html (http://people.hofstra.edu/Stefan_Waner/RealWorld/calctopic1/regression.html)
and i have been trying to do my question using example 3, (q = Ar^t is also totally new to me) im just having a go at it right now and will let you know how it goes, am i on the right path?? one thing im bit confused about right now is that the examples final answer is R = 2.6770 (1.3774)^t but mine is gonna be ^-t/T? how is it gonna differ from the example?

Ahh man... I hate my maths teacher lol.

thank you so much galactus for your fast respond by the way.

Farvahar021
06-18-2011, 05:57 PM
Aright so i worked it out using example 3 on that page and i got
i=0.0199*0.00673^(-t/T)
my I in the equation equals to the same answer as galactus got, which you have no idea how happy it made me! but how do i find out what my big T is? how did you get 1/5?

mmm4444bot
06-18-2011, 07:02 PM
I got I=.02e^{\frac{-t}{\frac{1}{5}}}=.02e^{-5t}

I think that galactus meant to type a lower-case i, above.

In other words, I think he means:

I = 0.02

such that

i = 0.02 e^{-5t}

mmm4444bot
06-18-2011, 07:06 PM
i got

i = 0.0199 * 0.00673^(-t/T)

Oops. Did you make a typographical error, too?

e does not equal 0.00673

mmm4444bot
06-18-2011, 07:09 PM
I did not use a regression method.

For part (a), I substituted given values of i and t into the formula, to create systems of two equations.

Each of the five systems that I solved yielded I = 0.02 and T = 0.20 (both rounded).

Farvahar021
06-18-2011, 07:28 PM
i got

i = 0.0199 * 0.00673^(-t/T)

Oops. Did you make a typographical error, too?

e does not equal 0.00673

so you just leave e as it is? because i though you need an answer for e therefore you can use it to solve question b and c by substituting the given number in... :oops:

Farvahar021
06-18-2011, 07:30 PM
I did not use a regression method.

For part (a), I substituted given values of i and t into the formula, to create systems of two equations.

Each of the five systems that I solved yielded I = 0.02 and T = 0.20 (both rounded).

could you please give me an example of one of your systems that you worked out? i dont understand how you worked out those to numbers by subtituting the numbers in the formula and how u created 2 equations...

by any chance does T equal 0.20 because its the difference in t? or is that just a dumb question :S

mmm4444bot
06-18-2011, 07:33 PM
i [thought] you need [a value] for e [before] you can use [the formula] to solve [questions] b and c

You are correct.

The value of e (rounded to five decimal places) is 2.71828

From where did the value 0.00673 come?

Why did you substitute 0.00673 for e?

Farvahar021
06-18-2011, 07:46 PM
i [thought] you need [a value] for e [before] you can use [the formula] to solve [questions] b and c

You are correct.

The value of e (rounded to five decimal places) is 2.71828

From where did the value 0.00673 come?

Why did you substitute 0.00673 for e?

Well using the example 3 from the page i mentioned before it asked me to log the values of i and put my number into the "on-line regression utility" after i entered my numbers into that table i got an equation of y=-2.172x + -1.699 in form of y = mx + b.
I was also told that in R = Ar^t r=10^m and A=10^b (im assuming A equal e?) and when i work out A it gives me 0.00673, where did i go wrong :S

mmm4444bot
06-18-2011, 08:21 PM
where did i go wrong

I'm not sure because I do not have time to review those other pages, but it seems to me that you tried to calculate the value of e when you should already have it memorized.

Shame on your instructor, if they handed out this exercise without first assuring that the class knows about e.

The symbol e represents a famous number.

This number is the base of the natural logarithm. Some people refer to it as Euler's Constant (pronounced "oilers"), after the Swiss mathematician.

e is an irrational number; its decimal form continues forever:

e = 2.718281828459045…

Math students need to know about e before they study exponential models, like the formula in your exercise.

Yes — I will type out one of the systems of equations that I formed and solved. Please give me a few minutes. 8-)

mmm4444bot
06-18-2011, 08:56 PM
i = I e^{-t/T}

From the chart, we have i = 0.0156 when t = 0.05

Substituting these two values into the exponential model gives us equation #1.

0.0156 = I e^{-0.05/T}

Likewise, substituting i = 0.0212 when t = 0.1 gives us equation #2.

0.0121 = I e^{-0.1/T}

Now, there are different methods to solve this system of two equations for I and T.

I began with these two steps on each equation:

(1) Divide both sides by I

(2) Take the natural logarithm of each side

These steps yield the following equations.

ln(0.0156/I) = \frac{-0.05}{T}

ln(0.0121/I) = \frac{-0.1}{T}

Next, I applied a property of logarithms to the lefthand side of each of these results.

ln(0.0156) - ln(I) = \frac{-0.05}{T}

ln(0.0121) - ln(I) = \frac{-0.1}{T}

Since the term ln(I) appears in each equation, I eliminated it from the system by subtraction. In other words, I subtracted the second equation above from the first:

[ln(0.0156) - ln(I)] - [ln(0.0121) - ln(I)] = \frac{-0.05}{T} - \frac{-0.1}{T}

ln(0.0156) - ln(0.0121) - ln(I) + ln(I) = \frac{-0.05 - (-0.1)}{T}

ln(0.0156) - ln(0.0121) = \frac{0.05}{T}

I used a scientific calculator to evaluate the lefthand side above. I then solved for T.

0.2541 = \frac{0.05}{T}

T = \frac{0.05}{0.2541} = 0.1968

Knowing the value of T allowed me to calculate the value of I, using either equation #1 or equation #2.

Substituting 0.1968 for T into equation #2 yields:

0.0121 = I e^{-0.1/0.1968}

0.0121 = I e^{-0.5081}

Again, using a scientific calculator, I evaluated the power of e above. I then solved for I.

0.0121 = 0.6016 I

I = \frac{0.0121}{0.6016} = 0.0201

From the given chart, I formed five pairs of equations (i.e., systems) from the given values of i and t.

As long as the results for I and T are rounded to two places, they always turn out to be 0.02 and 0.2, respectively.

This approach allows one to determine the answers in this exercise, but it has nothing to do with "linearizing using logs". Maybe that is associated with the regression method. What topic is your class currently covering?

Use the model i = 0.02 e^{-t/0.2} to answer parts (b) and (c).

Hmmm, I just noticed that part (b) asks for I instead of i, when t = 12. I'm not sure why. Maybe it's a typo?

Also, 12 milliseconds is so far outside the data in the chart that our values of I and T might not apply because of bad extrapolation. Maybe it's supposed to be 1.2? :?

Farvahar021
06-19-2011, 08:52 AM
Yes I get it now and I just done all a, b and c. Thank you so so much for taking your time out and helping me out!! :D

for b) I just pretended that its an error and should be i instead of capital I, because from what I understand I is a constant which we found in a) and wouldn't change right?

Again thank you very much mmm4444bot and also galactus.