Notes - Logarithmic Properties Def: Logarithms are inverses of exponential functions. y = logbx y = bx Therefore, y = logbx is equivalent to x = by logarithmic form Ex1 exponential form Write each equation in exponential form: a) 2 = log5x b) 3 = logb64 y = logbx → by = x y = logbx → by = x 52 = x Ex2 b3 = 64 c) log37 = y y = logbx → by = x d) logba = r y = logbx → by = x 3y = 7 br = a Express each in logarithmic form: a) 122 = x by = x → y = logbx log12x = 2 b) b3 = 8 by = x → y = logbx logb8 = 3 c) ey = 9 by = x → y = logbx d) rm = v by = x → y = logbx loge9 = y logrv = m On a separate sheet of paper to turn in at the end of class: Express each in exponential form: Express each in exponential form: 1) 4 = log216 1) 4 = log216 → 24 = 16 2) 2 = log9x 2) 2 = log9x → 92 = x 3) 3 = logb27 3) 3 = logb27 → b3 = 27 4) log5125 = y 4) log5125 = y → 5y = 125 Express each in logarithmic form: Express each in logarithmic form: 5) 23 = 8 5) 23 = 8 → log28 = 3 2 6) 15 = x 6) 152 = x → log15x = 2 7) b3 = 1000 7) b3 = 1000 → logb1000 = 3 y 8) 300 = 8 8) 300 = 8y → log8300 = y Properties of Logarithms In general, with any base b: logbb = 1 logb1 = 0 logbbx = x blog x = x Def: Natural Log : lne = 1 ln1 = 0 lnex = x elnx = x Common Log: log10 = 1 log1 = 0 log10x = x 10logx = x The common log is log base 10. We don't write the base 10 (log10), we simply write log. It is understood. On a separate sheet of paper to turn in at the end of class: Simplify each using the 15) 7log723 Simplify each using the properties of logarithms. 16) log108 properties of logarithms. log33 9) log1111 17) 10 9) log1111 = 1 10) log41 18) ln1 10) log41 = 0 11) log61 19) lne 11) log61 = 0 7 6 12) log55 20) lne 12) log557 = 7 6 ln125 13) log44 21) e 13) log446 = 6 log 19 9x 14) 8 8 22) lne 14) 8log819 = 19 15) 7log723 = 23 16) log108 = 8 17) 10log33 = 33 18) ln1 = 0 19) lne = 1 20) lne6 = 6 21) eln125 = 125 22) lne9x = 9x What if you are asked to simplify a logarithm that is not expressed as a property? Ex3 Evaluate log39 think 3 to the what power is 9? → well, 32 = 9, so log39 = 2 Evaluate log5125 think 5 to the what power is 125? → well, 53 = 125, so log5125 = 3 Evaluate log164 think 16 to the what power is 4? → well, 16 = 4, and 16 = 16!/! , so 16!/! = 4 therefore, log164 = ½ On a separate sheet of paper to turn in at the end of class: Evaluate each: Evaluate each: 23) log216 23) log216 = 4 24) log100 24) log100 = 2 25) log366 25) log366 = 1/2 26) log749 26) log749 = 2 27) log327 27) log327 = 3 28) log264 28) log264 = 6 Ex4 Evaluate log5140 think 5 to the what power is 140? → well, 53 = 125 and 54 = 625, hmm… so the answer must be a decimal before 3 and 4, but closer to 3. Exactly what decimal? The Change of Base Property !"# (!) !" (!) 𝑙𝑜𝑔! 𝑀 = !"# (!) 𝑎𝑛𝑑 𝑙𝑜𝑔! 𝑀 = !" (!) Ex4 Evaluate log5140 I know that the answer must be between 3 and 4. Use the change of base formula: !"# (!"#) 𝑙𝑜𝑔! 140 = !"# (!) = 3.07 Ex5 Evaluate log753 !" (!") 𝑙𝑜𝑔! 53 = !" (!) = 2.04

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