TutorState.com: TutorState.com - Logarithms

TutorState.com: TutorState.com - Logarithms

TutorState.com - Logarithms

TutorState.com - Logarithms
http://universitytutoringservice.heyclick.net/content/sid/1/page/logs.html

Logarithms

Introduction

Logarithms were invented by Napier. Before we had calculators, logarithms made calculation easier because they reduced multiplication and divisions to addition and subtraction. These days, logarithms are less important for this purpose.

Logarithms

A logarithm is writen as log_a(x) where a is a number called the base. Usually logarithms are written to the base 10 or sometimes base 2 for binary numbers, but it can be any number. If the logarithm is to the base e= 2.71828..., then we call it a natural logarithm because it is the only logarithm base which has a rate of change equal to the thing which is changing. Natural logs are also written, ln(u). The ln comes from the Latin, log naturalis.

Rules of Logarithms

log_a(0) is undefined         (1)
log_a(1) = 0                          (2)
log_a(u v) = log_a(u) + log_a(v)        (3)
log_a(u/v) = log_a(u) - log_a(v)         (4)
log_a(u)ⁿ = n log_a(u)                      (5)
log_b(u) = log_a(u)/log_a(b) - change of base      (6)
Where b is the old base, a is the new base, u is the argument of the logarithm.
log_a(1/u) = - log_a(u) from (2) and (4).                (7)
With these rules we can manipulate the exponential functions. Logs are the inverse functions of

Exponential Function

We have seen how to convert a number into a logarithmic number but how about if we are given a logarithm of a number and want to know what the original number was? This is achieved using exponential functions. The exponential function is the inverse function of a logarithmic function.
That is to say, that a^log_a(u) = u or log_a(a^u)=u
If we have a log to the base 10, the inverse function is 10^u.
For natural logarithms, the inverse function is e^u or exp(u). In particular, log₁₀(10^u) = u
and 10^log₁₀(u)= u              (8)
ln(e^u) = u or e^ln(u) = u     (9)

Example

What is the value of the number the gives the following logs to the base 10:
i) 1, ii) 12, iii) -2, iv) 2.6, v) -5.43

Answers

1. log₁₀ x = 1
   x=10¹ = 10
2. log₁₀ x = 12
   x=10¹² = 1x10¹²
3. log₁₀ x = -2
   x=10^-2 = 0.02
4. log₁₀ x = 2.6
   x=10^2.6 = 501.187
5. log₁₀ x = 12
   x=10^-5.43 = 3.71535 x 10^-6

Logarithmic Scales

Logarithms are used to increase the range over which numbers can be seen in a meaningful way. Logarithms are particularly useful when the data extends from the very small to the very large. There are many examples of there use:

• Astronomy - the magnitudes of stars
• In chemistry, the pH scale is a logarithmic scale which extends over 15 orders of magnetude, measuring the concentration of H ions in a solution. pH 0 is 10,000,000, De-ionised water is pH 7, 1 and pH 14 is 1/10,000,000
• Acoustics - the intensity of sound is measure in decibels (dB). dB= 10 log₁₀(I/I₀) where I is the intensity of the sound measured in W m^-2, I₀ is defined as 1x10^-12 W m^-2.
• Seismology - the Richter scale. An earthquake that measures 5 on the scale is ten times as powerful as a magnitude 4 earthquake, which is ten times as powerful as a magnitude 3 earthquake and so on.
• Engineers often use a logarithmic scale (dBm) to measure the output power of photonic devices.
• The Palermo Technical Impact Hazard Scale - a log scale that measures the probability of the Earth being struck by a meteorite of size. a Palermo Scale value of -2 indicates that the detected potential impact event is only 1% as likely as a random background event occurring in the intervening years, a value of zero indicates that the single event is just as threatening as the background hazard, and a value of +2 indicates an event that is 100 times more likely than a background impact by an object at least as large before the date of the potential impact in question.^[1]
• Casualties of War^[2] A log scale to measure the total number of deaths in different conflicts. A single death would correspond to a magnitude 0. A terrorist campaign that kills 100 people would correspond to a magnitude of 2. The second world war had a total of 62,000,000 casualties, which would correspond to a magnitude 7.79. The ratio of the number of deaths is easy to compare, for example in the Viet-Nam conflict, the US lost about 50,000 soldiers, (magnitude 4.69), in the civil war of Independence, the US losses were 6824 (magnitude 3.83). The ratio of casualties is therefroe, 4.69-3.83 = 0.864 which is 10^0.864 ≅ 7.3 times.

  The Lottery of Life

  It has even been mooted that we should adopt a logarithmic risk scale. The disadvantage of this, however, is that relatively few people understand logarithms well enough not be confused. However, it has been suggested^[3], that people already understand the 'risk' of winning different prizes from the National Lottery. The probability of each greater prize roughly increases logarithmically. The probability of an event is matched with corresponding likelyhood of drawing so many balls, given a £5 stake. This could be particularly useful for medical practictioners, who have to convey a wide range of risks to the general public. The logarithmic risk is given by,
  10 - log₁₀(1/P(x)) where P(x) is the probability of the event occuring. Table 1. sets out the probability of winning Lottery scales, probability, verbal risk scale and log risk scale.
  Using this scale, a highly probable event is comparable to drawing 3 matching balls, or a 9 on the log scale. A moderately likely event compares to the chance of drawing 4 matching balls, or 8 on the log scale and so on. A highly unlikely event compares to drawing 6 balls, which carries a 'risk' of 1 in 2,796,763. (£5 stake remember) or less than 4 on the log scale. In one year, assuming you enter the 2 draws per week, this is 104 draws a year. Therefore, the chance of winning in one year on a £5 stake is 104/2796763 = 1 in 26892. or 5.57 on the log scale.
  

  No. of Balls	Probability P(x)	Verbal scale	Log Scale
  3	1 in 11	High	9
  4	1:206	Moderate	8
  4+B	1 in 8878	Low	7
  5	1 in 11 098	Very low	6
  5+B	1 in 466,127	Minimal	4-5
  6	1 in 2,796,763	Negligible	<4
  B = bonus ball

  Table 1. The risk of winning the UK National Lottery.

  Another unlikely event is being murdered. It is a difficult risk to assign. It depends on many factors, such as your age, social-class, etc. In the UK, a figure of 51 in 1,000,000 has been suggested as a the risk of being murdered in a year, which equates to a probability of 1 in 19,600. (5.7 log scale)
  People place bias on good events over bad events (or bad events over good, if you work in the media) which distort their ideas of risk. I meet people that play the National Lottery and believe that they stand a good chance of winning the jackpot but the same people do not go out thinking, 'Ha-ha. This could be my lucky-day!' about being murdered so I have my doubts about the usefulness of this log scale.
   

References

[1] "Quantifying the risk posed by potential Earth impacts" by Chesley et al. (Icarus 159, 423-432 (2002)).

[2] Brian Hayes, “Statistics of deadly quarrels,”Am. Sci.90, 10–13 (Jan.–Feb. 2002).

[3] D M Campbell, "Risk language and dialects Expressing risk in relative rather than absolute terms is important", BMJ. 1998 April 18; 316(7139): 1242.

Logarithms

Logarithms

Chemistry Tutor - Private Tutor for Chemistry in Queens and NYC Metro Area

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Educational Tutoring: K-12 Math, Science, SAT

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Problem Solving

Problem Solving: Draw a Picture

Problem Solving: Draw a Picture

What Is It?

The draw a picture strategy is a problem-solving technique in which students make a visual representation of the problem. For example, the following problem could be solved by drawing a picture:

A frog is at the bottom of a 10-meter well. Each day he climbs up 3 meters. Each night he slides down 1 meter. On what day will he reach the top of the well and escape?

Why Is It Important?

Drawing a diagram or other type of visual representation is often a good starting point for solving all kinds of word problems. It is an intermediate step between language-as-text and the symbolic language of mathematics. By representing units of measurement and other objects visually, students can begin to think about the problem mathematically. Pictures and diagrams are also good ways of describing solutions to problems; therefore they are an important part of mathematical communication.

How Can You Make It Happen?

Encourage students to draw pictures of problems at the very beginning of their mathematical education. Promote and reinforce the strategy at all subsequent stages. Most students will naturally draw pictures if given the slightest encouragement.
Introduce a problem to students that will require them to draw a picture to solve. For example:

Marah is putting up a tent for a family reunion. The tent is 16 feet by 5 feet. Each 4-foot section of tent needs a post except the sides that are 5 feet. How many posts will she need?

Demonstrate that the first step to solving the problem is understanding it. This involves finding the key pieces of information needed to figure out the answer. This may require students reading the problem several times or putting the problem into their own words.

16 feet by 5 feet
1 post every 4 feet, including 1 at each corner
No posts on the short sides

1. Choose a Strategy
   Most often, students use the draw a picture strategy to solve problems involving space or organization, but it can be applied to almost all math problems. Also students use this strategy when working with new concepts such as equivalent fractions or the basic operations of multiplication and division.
    

2. Solve the Problem
   Students understand that there are posts every 4 feet. In the second sample problem, students are asked to organize data spatially to determine the number of posts Marah will need. They can draw a picture or a diagram to find the answer.
    

   I drew a rectangle where each long side is 16 feet, and there is 1 post every 4 feet. I drew a circle for each post. I remembered to draw a post at each end. There are 10 posts total.

3. Check Your Answer
   Ask students to read the problem again to be sure they answered the question.
   

   I found that there are 10 posts.

   Students should check their math to be sure it is correct.
   

   16 divided by 4 is 4. There are 4 sections of 4 feet on each long side.
   There is a post on each end, so 4 + 1 = 5. There are 2 sides to the tent, and 5 x 2 = 10.

   Discuss with students whether draw a picture was the best strategy for this problem. Was there a better way to solve it?
   Drawing a picture was a good strategy to use for this problem because students might forget to count the posts on each corner unless they see them.
    
4. Explain How You Found the Answer
   Students should explain their answer and the process they went through to solve the problem. It is important for students to talk or write about their thinking. There may be more than one way to represent a problem visually, and asking students to explain their picture helps to understand their thinking process and identify errors.
   

   My answer is 10 fence posts. First, I tried to solve this by multiplying. I took 16 and divided by 4 to find the number of posts on each side. I got 4 posts on each side. Then I doubled it to get 8 posts total. I checked the problem and realized that there are posts on each corner, so I drew a picture so that I could see it and be sure the answer was correct.

   I drew a rectangle to show the tent. Each long side is 16 feet, and there is 1 post every 4 feet, so I divided 16 by 4 to find out that there are 4 sections of 4 feet each. I drew a circle for each post, and wrote the number in the space between each post. I remembered to draw a post at each end. I counted the posts and found out that there are 10 posts total.

5. Guided Practice
   Have students try to solve the following problem using the draw a picture strategy.
   

   Tai wants to frame a 3 x 5 picture surrounded by 2 inches of mat. How large will her frame need to be?

   Have students work in pairs, groups, or individually to solve this problem. They should be able to tell or write about how they found the answer as well as be able to justify their reasoning.
    

How Can You Stretch Students' Thinking?

Some students are visual learners and work well when problems are illustrated or easy to see. Encourage students to draw pictures or diagrams for problems they find difficult. Encourage students to label all parts of their drawings. Students should understand that their drawings do not need to be perfect. Rather, their drawings need only represent the problem accurately and clearly show their thought processes.


http://www.teachervision.fen.com/math/problem-solving/48931.html
Read more on TeacherVision: http://www.teachervision.fen.com/math/problem-solving/48931.html?page=2&detoured=1#ixzz14vFGpIfH

integrate 1/(sinx + cosx) dx

integrate 1/(sinx + cosx) dx
maybenow.com
∫ [1 /(cosx + sinx)] dx = multiply and divide the integrand by (cosx - sinx): ∫ {(cosx - sinx) /[(cosx + sinx)(cosx - sinx)]} dx = expand the denominator: ∫ [(cosx - sinx) /(cos²x - sin²x)] dx = break it up into: ∫ [cosx /(cos²x - sin²x)] dx + ∫ [- sinx /(cos²x - sin²x)] dx = rewrite the first denominator in terms of sinx and the second one in terms of cosx: ∫ {cosx /[(1 - sin²x) - sin²x]} dx + ∫ {- sinx /[cos²x - (1 - cos²x)]} dx = ∫ [cosx /(1 - sin²x - sin²x)] dx + ∫ [- sinx /(cos²x - 1 + cos²x)] dx = ∫ [cosx /(1 - 2sin²x)] dx + ∫ [- sinx /(2cos²x - 1)] dx (#) let us solve the first integral substituting sinx = t hence (differentiating both sides) d(sinx) = dt → cosx dx = dt, yielding: ∫ cosx dx /(1 - 2sin²x) = ∫ dt /(1 - 2t²) = factor the denominator as a difference of squares: ∫ dt /{1 - [(√2)t]²} = ∫ dt /{[1 - (√2)t][1 + (√2)t]} = decompose it into partial fractions: 1 /{[1 - (√2)t][1 + (√2)t]} = A/[1 - (√2)t] + B/[1 + (√2)t] 1 /{[1 - (√2)t][1 + (√2)t]} = {A[1 + (√2)t] + B[1 - (√2)t]} /{[1 - (√2)t][1 + (√2)t]} 1 = A + (√2)At + B - (√2)Bt 1 = (√2)(A - B)t + (A + B) hence: | (√2)(A - B) = 0 | A + B = 1 | A = B | B + B = 1 | A = 1/2 | B = 1/2 yielding: 1 /{[1 - (√2)t][1 + (√2)t]} = A/[1 - (√2)t] + B/[1 + (√2)t] = (1/2)/[1 - (√2)t] + (1/2)/[1 + (√2)t] thus the integral becomes: ∫ [1 /(1 - 2t²)+ dt = ∫ {{(1/2)/[1 - (√2)t]} + {(1/2)/[1 + (√2)t]}} dt = break it up pulling constants out: (1/2) ∫ {1 /[1 - (√2)t]} dt + (1/2) ∫ {1 /[1 + (√2)t]} dt = divide and multiply the first integral by (-√2), and the second one by (√2) so as to make each numerator the derivative of the respective denominator: (1/2)(-1/√2) ∫ {(-√2) /[1 - (√2)t]} dt + (1/2)(1/√2) ∫ {(√2) /[1 + (√2)t]} dt = [- 1/(2√2)] ∫ d[1 - (√2)t] /[1 - (√2)t]} + [1/(2√2)] ∫ d[1 + (√2)t] /[1 + (√2)t] = [- 1/(2√2)] ln |1 - (√2)t| + [1/(2√2)] ln |1 + (√2)t| + C = [1/(2√2)] [ln |1 + (√2)t| - ln |1 - (√2)t|] + C = recalling logarithms properties, [1/(2√2)] ln |[1 + (√2)t] /[1 - (√2)t]| + C thus, substituiting back sinx for t, you have: ∫ [cosx /(1 - 2sin²x)] dx = [1/(2√2)] ln |[1 + (√2)sinx] /[1 - (√2)sinx]| + C (##) the solution of the latter integral (see expression (#) above) is similar: ∫ [- sinx /(2cos²x - 1)] dx = let cosx = u → d(cosx) = du → - sinx dx = du ∫ [- sinx /(2cos²x - 1)] dx = ∫ du /(2u² - 1) = ∫ du /[(√2)u]² - 1] = ∫ du /{[(√2)u - 1][(√2)u + 1]} partial fraction decomposition 1 /{[(√2)u - 1][(√2)u + 1]} = A/[(√2)u - 1] + B/[(√2)u + 1] 1 /{[(√2)u - 1][(√2)u + 1]} = {A[(√2)u + 1] + B[(√2)u - 1]} /{[(√2)u - 1][(√2)u + 1]} 1 = A[(√2)u + 1] + B[(√2)u - 1] 1 = (√2)Au + A + (√2)Bu - B 1 = (√2)(A + B)u + (A - B) | (√2)(A + B) = 0 | A - B = 1 | A = - B | - B - B = 1 | A = 1/2 | B = - 1/2 yielding: 1 /{[(√2)u - 1][(√2)u + 1]} = A/[(√2)u - 1] + B/[(√2)u + 1] = (1/2)/[(√2)u - 1] - (1/2)/[(√2)u + 1] hence: ∫ du /(2u² - 1) = ∫ {{(1/2)/[(√2)u - 1]} - {(1/2)/[(√2)u + 1]}} du = (1/2) ∫ du /[(√2)u - 1] - (1/2) ∫ du /[(√2)u + 1] = dividing and multiplying by √2, (1/2)(1/√2) ∫ (√2) du /[(√2)u - 1] - (1/2)(1/√2) ∫ (√2) du /[(√2)u + 1] = [1/(2√2)] ∫ d[(√2)u - 1] /[(√2)u - 1] - [1/(2√2)] ∫ d[(√2)u + 1] /[(√2)u + 1] = [1/(2√2)] ln |(√2)u - 1| - [1/(2√2)] ln |(√2)u + 1| + C = [1/(2√2)] [ln |(√2)u - 1| - ln |(√2)u + 1|] + C = [1/(2√2)] ln |[(√2)u - 1] /[(√2)u + 1]| + C = substituiting back cosx for u, ∫ [- sinx /(2cos²x - 1)] dx = [1/(2√2)] ln |[(√2)cosx - 1] /[(√2)cosx + 1]| + C (###) thus, plugging this and the previous (###) result into the above (#) expression, you have: ∫ [cosx /(1 - 2sin²x)] dx + ∫ [- sinx /(2cos²x - 1)] dx = [1/(2√2)] ln |[1 + (√2)sinx] /[1 - (√2)sinx]| + [1/(2√2)] ln |[(√2)cosx - 1] /[(√2)cosx + 1]| + C = [1/(2√2)] {ln |[1 + (√2)sinx] /[1 - (√2)sinx]| + ln |[(√2)cosx - 1] /[(√2)cosx + 1]|} + C = owing to logarithm properties, [1/(2√2)] ln |{[1 + (√2)sinx] /[1 - (√2)sinx]}{[(√2)cosx - 1] /[(√2)cosx + 1]|} + C thus, in conclusion: ∫ [1 /(cosx + sinx)] dx = [1/(2√2)] ln |{[1 + (√2)sinx][(√2)cosx - 1]} /{[1 - (√2)sinx][(√2)cosx + 1]}| + C