2.05 ExpLog - Logarithms Intro | High School

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2.05 INTRO TO LOGARITHMS LOGARITHMS A logarithm is another way to write an exponential expression. 𝟐 𝟏𝟎 = 𝟏𝟎𝟎 𝒍𝒐𝒈𝟏𝟎 𝟏𝟎𝟎 = 𝟐 LABELING EXPRESSIONS Exponential form Logarithmic form 𝟏𝟎𝟐 = 𝟏𝟎𝟎 𝒍𝒐𝒈𝟏𝟎 𝟏𝟎𝟎 = 𝟐 10 = base (B) 2 = exponent (E) 100 = power (P) SWITCHING FORMS 𝑬 𝑩 =𝑷 𝒍𝒐𝒈𝑩 𝑷 = 𝑬 𝑩𝑬 = 𝑷 TRY IT 𝟓𝟑 = 𝟏𝟐𝟓 𝒙𝟐 = 𝟐𝟓 𝒍𝒐𝒈𝑩 𝑷 = 𝑬 𝟏 𝟐 𝟑 𝟏 = 𝟖 𝟑𝟒 = 𝒙 CALCULATOR ln log This means log 𝟏𝟎 ( 𝐰𝐡𝐚𝐭𝐞𝐯𝐞𝐫 𝐲𝐨𝐮 𝐭𝐲𝐩𝐞) 𝟏𝟎𝒂𝒏𝒔𝒘𝒆𝒓 = (𝒘𝒉𝒂𝒕𝒆𝒗𝒆𝒓 𝒚𝒐𝒖 𝒕𝒚𝒑𝒆) SOME: log 𝒆 ( 𝐰𝐡𝐚𝐭𝐞𝐯𝐞𝐫 𝐲𝐨𝐮 𝐭𝐲𝐩𝐞) 𝒆𝒂𝒏𝒔𝒘𝒆𝒓 = (𝒘𝒉𝒂𝒕𝒆𝒗𝒆𝒓 𝒚𝒐𝒖 𝒕𝒚𝒑𝒆) log ∎ ∎ Lucky You can do ALL bases PROPERTIES OF LOGARITHMS 1) Change of Base Law 𝒍𝒐𝒈𝒔 𝒎 = 𝒍𝒐𝒈𝒄 𝒎 𝒍𝒐𝒈𝒔 𝒄 Ex: 𝒍𝒐𝒈𝟗 𝟖 𝒍𝒐𝒈𝟗 𝟐 = 𝒍𝒐𝒈𝟐 𝟖 PROPERTIES OF LOGARITHMS Apply it to solve with some calculators 𝒍𝒐𝒈𝟏𝟎 𝒎 𝒍𝒐𝒈𝒄 𝒎 = 𝒍𝒐𝒈𝟏𝟎 𝒄 Ex: 𝒍𝒐𝒈𝟐 𝟖 = 𝒍𝒐𝒈𝟏𝟎 𝟖 𝒍𝒐𝒈𝟏𝟎 𝟐 Calculator Fix!! PROPERTIES OF LOGARITHMS 2) Fundamental Property 𝒍𝒐𝒈 𝒎 𝒄 𝒄 Ex: 𝒍𝒐𝒈 𝟖 𝟐 𝟐 =𝒎 =𝟖 1. Rewrite as exponential equation 2. Is this statement true? PROPERTIES OF LOGARITHMS 3) Product Law 𝒍𝒐𝒈𝒄 𝒎𝒏 = 𝒍𝒐𝒈𝒄 𝒎 + 𝒍𝒐𝒈𝒄 𝒏 32 = ? 4 ×8 Ex: 𝒍𝒐𝒈𝟐 𝟑𝟐 = 𝒍𝒐𝒈𝟐 𝟒 + 𝒍𝒐𝒈𝟐 𝟖 2? = 32 =𝟓 2? = 4 =𝟐+𝟑 =𝟓 2? = 8 PROPERTIES OF LOGARITHMS Product Law – The other way 𝒍𝒐𝒈𝒄 𝒎 + 𝒍𝒐𝒈𝒄 𝒏 = 𝒍𝒐𝒈𝒄 𝒎𝒏 Ex: 𝒍𝒐𝒈𝟑 𝟑𝒙 + 𝒍𝒐𝒈𝟑 𝒙 + 𝒍𝒐𝒈𝟑 𝟔𝒚 𝑎𝑟𝑒 𝑡ℎ𝑒𝑦 𝑎𝑙𝑙 𝑠𝑎𝑚𝑒 𝑏𝑎𝑠𝑒? Combine the powers 𝒑𝒐𝒘𝒆𝒓 = (𝟑𝒙)(𝒙)(𝟔𝒚) = 𝟏𝟖𝒙𝟐 𝒚 = 𝒍𝒐𝒈𝟑 𝟏𝟖𝒙𝟐 𝒚 PROPERTIES OF LOGARITHMS 4) Quotient Law 𝒍𝒐𝒈𝒄 𝒎 = 𝒍𝒐𝒈𝒄 𝒎 − 𝒍𝒐𝒈𝒄 𝒏 𝒏 64 = ? 128 ÷ 2 Ex: 𝒍𝒐𝒈𝟐 𝟔𝟒 = 𝒍𝒐𝒈𝟐 𝟏𝟐𝟖 − 𝒍𝒐𝒈𝟐 𝟐 2? = 64 =𝟔 2? = 128 =𝟕−𝟏 =𝟔 2? = 2 PROPERTIES OF LOGARITHMS Quotient Law – The other way 𝒍𝒐𝒈𝒄 𝒎 − 𝒍𝒐𝒈𝒄 𝒏 = 𝒍𝒐𝒈𝒄 𝒎 𝒏 Ex: 𝒍𝒐𝒈𝟑 𝟏𝟖𝒚𝟐 − 𝒍𝒐𝒈𝟑 𝟑𝒚 𝑎𝑟𝑒 𝑡ℎ𝑒𝑦 𝑎𝑙𝑙 𝑠𝑎𝑚𝑒 𝑏𝑎𝑠𝑒? Combine the powers 𝒑𝒐𝒘𝒆𝒓 = 𝟏𝟖𝒚𝟐 ÷ (𝟑𝒚) = 𝟔𝒚 = 𝒍𝒐𝒈𝟑 𝟔𝒚 PROPERTIES OF LOGARITHMS 5) Power with Exponent Law 𝒍𝒐𝒈𝒄 Ex: 𝒍𝒐𝒈𝟐 𝒏 𝒎 𝟑 𝟒 43 = 64 = 𝒍𝒐𝒈𝟐 𝟔𝟒 =𝟔 = 𝒏𝒍𝒐𝒈𝒄 𝒎 By the Law = 𝟑𝒍𝒐𝒈𝟐 𝟒 2? = 4 ?= 2 = 𝟑(𝟐) =𝟔 PROCESSES Manipulating Log Expressions Solving with Logs Inverse – will come later MANIPULATING LOG EXPRESSIONS Simplify to ONE logarithm 𝟑𝒍𝒐𝒈 𝒙 + 𝟐𝒍𝒐𝒈 𝒚 MANIPULATING LOG EXPRESSIONS Simplify to ONE logarithm 𝟐𝒍𝒐𝒈𝟕 (𝒙) + 𝟓𝒍𝒐𝒈𝟕 (𝒚𝟒 ) − 𝟑𝒍𝒐𝒈𝟕 (𝟐) MANIPULATING LOG EXPRESSIONS Simplify to ONE logarithm 𝟒𝒍𝒐𝒈𝟓 𝒂𝒄 − 𝟑𝒍𝒐𝒈𝟓 (𝒂𝒃𝟐 ) − 𝟐𝒍𝒐𝒈𝟓 (𝒃𝒄) + 𝟐 MANIPULATING LOG EXPRESSIONS Expand **final solution has no exponents** 𝟑 𝒙+𝟒 𝒍𝒐𝒈𝒎 𝟐 𝟐 SOLVING WITH LOGS Solve 𝟒=𝟐 𝟖 𝒙−𝟒 −𝟖 Isolate the Exponent 𝟔= 𝟖 𝒙−𝟒 Switch Forms 𝒙 − 𝟒 = 𝒍𝒐𝒈𝟖 𝟔 Change of Base (if you need to) 𝒍𝒐𝒈𝟔 𝒙−𝟒= 𝒍𝒐𝒈𝟖 Solve 𝒙 − 𝟒 = 𝟎. 𝟖𝟔𝟏𝟔𝟓𝟒𝟏𝟔𝟔 𝒙 = 𝟒. 𝟖𝟔𝟏𝟔𝟓𝟒𝟏𝟔𝟔 𝑩𝑬 = 𝑷 𝒍𝒐𝒈𝑩 𝑷 = 𝑬 SOLVING WITH LOGS Solve 𝟒=𝟐 𝟕 𝟐𝒙+𝟑 − 𝟏𝟔 Isolate the Exponent 𝟏𝟎 = 𝟕 𝟐𝒙+𝟑 Switch Forms 𝟐𝒙 + 𝟑 = 𝒍𝒐𝒈𝟕 𝟏𝟎 Change of Base (if you need to) 𝒍𝒐𝒈𝟏𝟎 𝟐𝒙 + 𝟑 = 𝒍𝒐𝒈𝟕 Solve 𝟐𝒙 + 𝟑 = 𝟏. 𝟏𝟖𝟑𝟐𝟗𝟒𝟔𝟔𝟐 𝒙 ≈ −𝟎. 𝟗𝟎𝟖𝟒 𝑩𝑬 = 𝑷 𝒍𝒐𝒈𝑩 𝑷 = 𝑬 𝑩𝑬 = 𝑷 SOLVING WITH LOGS Try It: Solve 𝟏 𝟏𝟕 = 𝟑 𝟒 Isolate the Exponent Switch Forms Change of Base (if you need to) Solve 𝒙+𝟕 − 𝟏𝟏 𝒍𝒐𝒈𝑩 𝑷 = 𝑬