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Logarithms

Introduction to. Logarithms. Consider the statement. 1000 x 100 = 100 000. Remembering (hopefully!) that…. 10 3 = 1000 10 2 = 100 10 5 = 100 000. We can rewrite our original statement in power (index) format as. 10 3 x 10 2 = 10 5.

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Logarithms

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  1. Introduction to Logarithms

  2. Consider the statement 1000 x 100 = 100 000 Remembering (hopefully!) that…. 103 = 1000 102 = 100 105 = 100 000 We can rewrite our original statement in power (index) format as 103 x 102 = 105

  3. Our statement that 102 x 103 = 105 is just a specific case of the general rule am x an= am + n Power Rule 1 In Year 10, you learned two other rules which went hand-in-hand with Rule 1…. am ÷an= am – n Power Rule 2 (am )n= am n Power Rule 3 These last two rules can be easily verified using real numbers as we did on the previous slide. These three rules are the basis of all our logarithm work to come. MAKE SURE YOU KNOW THEM!!

  4. n Rule 7………a 1/n = a Help! I’m drowning in rules! There are also some other rules you need to remember as these appear in log work…. Rule 4………a 0 = 1 Rule 5………a 1 = a Rule 6………a -n = 1/a n

  5. Intro to Logs Now let’s take this a bit further and go beyond Year 10 work….. In the statement 100 = 102 2 is called the “logarithm”( known to Year 10s as “power” or “index” but in senior school we call it LOGARITHM !! 100 is called the “number” 10 is called the “base”, and SoLOGARITHM is just a fancy word forPOWER

  6. 100 = 102 SAME THING! log10 100 = 2 So, where are we….? can be described in words as “100 equals 10 squared”. BUT…using our new terminology from the earlier slide we can also say 2 is the LOGARITHM of the NUMBER 100 (BASE 10) and in symbols….

  7. This is called POWER FORMAT This is called LOGARITHMIC FORMAT So we now have these two interchangeable formats…… 100 = 102 log10100 = 2 Now, if we replace the 100, 10 and 2 with letters, we can come up with a formula which then enables us to do this interchange for all numbers, bases and logarithms

  8. 100 = 102 log10100 = 2 Replace 100 with n Replace 10 with a Replace 2 with t LEARN! n = at loga n = t Log Law #1 – the most important of all! This is known as the TRANSFORMATION RULE and must be memorised! It will enable you to swap between power format and log format with ease! Insect lovers take note! You might notice two insects, ANTs (which live in logs) and NATs (misspelt!).

  9. Exercises Use the Transformation Rule to fill in this table log39 = 2 81 = 34 log464 = 3 216 = 63 log2(1/4) = -2 1/125 = 5-3 log82 = 1/3 Click to check your answers

  10. By the Transformation Rule logan =t n = at log41024 =x 1024 = 4x Example 1 Find the value of log41024 Solution Let log4 1024 = x We can write Now solve this! It’s easier to solve power format than log format. Using calculator we trial various powers of 4, and ultimately we find that 45= 1024, sox = 5 There are more technically correct ways to do this, but for the moment, trial and error will do!

  11. By the Transformation Rule logan =t n = at log20.25 =x 0.25 = 2x Example 2 Evaluate without a calculator log 2 0.25 Solution Let log 2 0.25 = x We can write Now if we change 0.25 into ¼, which is 1/22 or 2-2, we then see that 2x = 2-2, and so x = – 2

  12. Remember our original statement on Slide 2 ? 1000 x 100 = 100 000 Which we then rewrote as 103 x 102 = 105 We can now go one better, and realising that the powers can be connected using 3 + 2 = 5, we can now write this another way, i.e. log101000 + log10100 = log10100 000

  13. log101000 + log10100 = log10100 000 Now if we replace 1000 with a 100 with b 100 000 with a x b 10 with n We can now write a general formula…… Log Law #2 log n a + log n b = log n (a x b) This is really just a disguised version of the Year 10 rule that when you multiply, you add the powers!

  14. Using a similar approach we can also show that log n a - log n b = log n (a ÷b) Log Law #3 which can also be written as log n a - log n b = log n (a /b)

  15. Example 3 Simplify log35 + log34 Solution Using Log Law #2 i.e. log n a + log n b = log n (ab ) Let n = 3, a = 5 and b = 4. This means ab = 20 So log35 + log34 = log320 (ans) Note: This is only possible when the base (n) is the same in both terms.

  16. Example 4 Simplify log272 – log29 Solution Using Log Law #3 i.e. log n a - log n b = log n (a / b ) Let n = 2, a = 72 and b = 9. This means a / b = 8 So log272 – log29 = log28 (ans – almost!) Now you should always check to see if these two numbers (the 2 and 8) are related in any way…… see next slide!

  17. Our answer is log28 but it can be simplified ! If you can get into the habit of checking if the 2 (base) and the 8 (number) are related as powers, you are then able to use the Transformation Rule…. log28 = x 8 = 2 x And so x = 3 A better ans then is: Note we could not have done this in Example 3 as 20 and 3 are not related. log272 – log29 = log28 = 3

  18. YAY!! Now for the last of the “Big Four” Remember Log Law #2 back on Slide 15? log n (ab) = log na + log nb This can be extended to more than two terms, e.g. log n (abc)= log na + log nb + log n c (3 terms) Or if a, b, c are all the same, say they’re all “ a ”…..then log n (a 3 )=log na + log na + log n a Check that you understand this before next slide! log n (a 3)= 3 log n a i.e.

  19. And if there are 4 terms, then…. log n (a 4)= 4 log n a And if there are “ y ” terms we can generalise to get our Log Law #4 Formula…. Log Law #4 log n (a y )= y log n a

  20. log a n = tn = at Summary so far.... LOG LAW #1: Transformation Rule LOG LAW #2: When numbers are multiplied, you ADD their logs log a (xy) = log ax + log ay LOG LAW #3: When numbers are divided, you SUBTRACT their logs log a (x / y) = log ax - log ay LOG LAW #4: The Power Law log a (x n) = n log ax

  21. In addition to the “Big 4”, there are also some “lesser” log laws which are special cases of the Big 4 and come in extremely handy! If you apply Law #1 (the Transformation Rule), you will see that a = a1 which is certainly true! Law #5: log aa = 1 Again applying the Transformation Rule gives 1 = a0 which is true! Law #6: log a1 = 0 Law #7: log a(1/x) = - logax Using Law #3 we first get log a 1 – log axand then using Law #6 above this becomes 0 – log a x i.e. – log a x

  22. Now we’ll do some examples which require all 7 log laws to be used strategically!

  23. Example 5 Simplify 2log35 + 3log34 Solution First use Law #4 to shift the coefficients (2 & 3) up to the power position 2log35 + 3log34 = log3 (52) + log 3 (43) which is log 3 25 + log 3 64 As this is now of the format log a + log b we can use Law #1 to combine together and get log (ab) = log3 (25x 64) At this stage, check if there is any recognisable power connection between 3 and 1600. Maybe check powers of 3 on the calc. There appears to be no connection, so leave this as the ans. Work it out = log3 (1600)

  24. Example 6 Simplify 2 log 5 3 - 2log 5 15 Solution Again use Law #4 to shift the coefficients (2 & 2) up to the power position 2log 5 3 - 2log 5 15 = log5 (32) - log 5 (152) which is log 5 9 - log 5 225 As this is now of the format log a - log b we can use Law #2 to combine together and get log (a / b) = log5 (9 / 225) At this stage, check if there is any recognisable power connection between 5 and 1/25. Now the 5 and 25 should give you the clue: 1/25 is equal to 5-2 !! OVER… Work it out = log5 (1/25)

  25. To work out log5 (1/25) Rewrite as log5 (5 -2 ) Use Law #4 to drop the power down the front = -2 log5 5 Now use Law #5 log a a = 1which can only be used then the base and number are the same!! (here they’re both 5!) = -2 x 1 Ans ! = -2

  26. Simplify Example 7 Solution Remember if you can spot a connection between the base (10) & numbers (100 & 1000), always work on this first, so rewrite the numbers as powers of 10. Now use Law #4 to drop powers (2 & 3) to the front Now use Law #5 Ans!

  27. If , find the value of a Example 8 Solution First use Law 3 to change the left side and kill the fraction Now use Law #6 loga1 = 0 Now remember 33 = 3 x 31/2 = 3 3/2 Now Law#4 to bring power down front which is now of form a log4 3 so a = -3/2

  28. Example 9 Express in simplest form 4 – 3log10x Solution This is the style of Q14, P283. The question is asking you to write 4 – 3 log 10x as a single log, i.e. in format log 10a. The overall strategy is firstly to write the “4” as log 10 (something) and use Law #4 to move the 3 to the power position. This will then give us a format log a – log bwhich we can then switch to log (a/b) using Law #3. PHEW!!! Here we go…

  29. Hmmmmmm…. what to do with the 4 ??? 4 – 3 log 10x Since there’s already a “log 10” present, maybe we could write the 4 as log 10 (something) ??

  30. log a n = tn = at log 10y = 4 y = 10 4 So Let 4 = log 10y Applying Law #1 (Transformation Rule) This means that y = 10 000 and so 4 = log10 10000 So back to the original question 4 – 3 log 10x can now be rewritten as log10 10000 – log 10x 3 using Law #4 to move the 3 = log10 (10000 /x 3) Ans !! using Law #2

  31. Example 10 Simplify 5log28 + 3 Solution Remember to first look for a connection between the 2 and 8 ? As 8 = 23 we can write log 2 8 = log 2 (23) 5log28 + 3 = 5log2 (23) + 3 Use Law #4 to move the 3 to front = 5 x 3 log 2 2 + 3 Use Law #5 to simplify log22 = 5 x 3 (1) + 3 =18 NOTE!! Here we didn’t have to change the “3” on the end into log (something), as we were able to simplify the first term and get rid of the log. This was possible because we made the effort to first find that connection between the 2 and the 8!

  32. Log Equations First revise: Transformation Rule Negative Indices Fractional Indices

  33. log a n = tn = at Example 11 Solve log 2x = 5 Solution Use LOG LAW #1: Transformation Rule log2 x = 5 so x = 25 Ans x = 32 Easy!!

  34. log a n = tn = at Example 12 Solve log 3 (1/9) = x Solution Use LOG LAW #1: Transformation Rule log3 (1/9)= x so 1/9 = 3x Ans x = -2 i.e. 3-2 = 3x Equating the powers, This is why I asked you to revise negative powers!!

  35. Example 13 Solve 3 x= 20 Note: This is a very common question where the unknown is in the power and there is no obvious connection between the two numbers (3 and 20 in this case). The strategy is to take log10 of BOTH SIDES then use LAW #4. 3 x= 20 First, take logs10 of both sides log10 (3x) = log1020 Now use Law#4 on left expression x log10 3 = log1020 Finally divide both sides by log10 3 to make x the subject Ans

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