Academic Skills Advice Logs A log has a base and a number: e.g. πππ3 8 base number This would be read as: βlog to the base 3 of 8β. The most common base is base 10. If you see a log without the base number (e.g. on your calculator) then you should assume it is base 10. e.g. πππ45 is usually read as βlog 45β and means βlog to the base 10 of 45β. What does a log mean? Examples: ππππ πππ means: βwhat power do I need to raise 5 to, to get 125?β β΄ πππ5 125 =3 (because 53 = 125) The answer to the log ππππ ππ means: β΄ πππ2 64 βwhat power do I need to raise 2 to, to get 64?β =6 ππππππ means: β΄ πππ100 © H Jackson 2011 / ACADEMIC SKILLS (because 26 = 64) βwhat power do I need to raise 10 to, to get 100?β =2 (because 102 = 100) 1 Change of base Most new calculators will accept a log to any base by using the general log key ( Enter base here ) Enter number here If you have an older calculator (without the general log key) and need to find a log with a different base then you can convert to base 10 using the formula: ππππ₯ π΄ = πππ10 π΄ πππ10 π₯ Remember to do: log of number divided by log of base. Example: ο· Find ππππ ππ using the log to the base 10 key. πππ4 35 = πππ10 35 πππ10 4 = π. ππ On your calculator you just need to type: πππ(ππ) ÷ πππ(π) Try finding the logs from the previous examples using your calculator (and either the general log key or the change of base formula). What about ln? Ln is a special log because itβs base is e (e is a number approximately equal to 2.718) Example: Ln 7 means βwhat power do I need to raise βeβ to, to get 7?β Ln 7 = 1.946 (3dp) because e1.946 = 7 Remember: e.g. e and ln βundoβ each other because they are βoppositesβ. ln(π 3 ) = 3 π ππ5 = 5 (because the ππ and the π undo each other (or cancel each other out)) (because the ππ and the π undo each other (or cancel each other out)) © H Jackson 2011 / ACADEMIC SKILLS 2 Laws of Logs There are a few rules/laws related to logs that it is useful to learn and know how to use. πππ(π¨π©) = πππ(π¨) + πππ(π©) π¨ πππ ( ) = π₯π¨π (π¨) β π₯π¨π (π©) π© πππ(π¨π ) = ππππ(π¨) πππ(π) = π (using ANY base) Examples: ο· Write π₯π¨π (π) + π₯π¨π (π) as a single log. Using the 1st rule: ο· Write π₯π¨π (ππ) β π₯π¨π (π) as a single log. Using the 2nd rule: ο· 12 log(12) β log(3) = log ( 3 ) = log(4) Find πππππ π. Using the 4th rule: ο· log(4) + log(5) = log(4 × 5) = log(20) πππ15 1 = 0 Write π π₯π¨π π (π) + ππ₯π¨π π (π) as a single log. We canβt just multiply the 5 and the 7 because of the numbers in front of the logs. We need to deal with them 1st. ο· Using the 3rd rule: 2 log 4 (5) + 3 log 4 (7) = log 4 (52 ) + log 4 (73 ) Using the 1st rule: log 4 (52 ) + log 4 (73 ) = log 4 (52 × 73 ) = log 4 (8575) Given that π₯π¨π (π) β π. πππ, find πππ(πππ) without using a calculator. You could use the 1st OR the 3rd rule for this question. Using the 1st rule: OR: Using the 3rd rule: © H Jackson 2011 / ACADEMIC SKILLS log(125) = log(5 × 5 × 5) = log(5) + log(5) + log(5) = 0.699 + 0.699 + 0.699 = π. πππ log(125) = log(53 ) = 3log(5) = 3 × 0.699 = π. πππ 3
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