with logs - Durham College

Student Academic Learning Services Page 1 of 2Exercises for Logs - SolutionsConverting log equations to exponential equationsConvert the following logarithmic equations into exponential equations, and solve for theunknown if possible.log 5 x = y log 2 x = 12 ln y = 4.918 log a = −1.42 ln x = ln 50x = 5 y x = 2 12x = 4096y = e 4.918y ≈ 136.7a = 10 −1.42a ≈ 0.38ln 50x = ex = 50Converting exponential equations to log equationsConvert the following exponential equations into logarithmic equations, and solve for theunknown if possible.10 x = 20 e y = 85 a c = b e −y = 0.452 2 x = 128x = log 20x ≈ 1.3y = ln 85y ≈ 4.44c = log a b −y = ln 0.452y = − ln 0.452y ≈ 0.794x = log 2 128x = 7Combining logs(Combining logs is important for solving log equations.)Express the following expressions using a single logarithm.3log 6 + log y log(7) − log (10) log √x= log (6y)+ 3log (x 2 )= log 710 = logx 1⁄ 3 + log(x 2 ) 3= logx 1⁄ 3 ∙ x 6 = log (x 2 )1log a − 2 log 63= log a 1⁄ 3 − log 6 2= log a1 ⁄ 36 2 3= log √a36 www.durhamcollege.ca/sals Student Services Building (SSB), Room 204905.721.2000 ext. 2491This document last updated: 7/5/2012

Student Academic Learning Services Page 2 of 2Solving log equationsHint: The key here is to combine all the logs into one log, and then rearrange it into anexponential equation before solving.5 log 2 x = 20log 2 x = 20 5log 2 x = 4x = 2 4x = 16log(8x 2 ) − log(2x) = 3log 8x22x = 3log(4x) = 34x = 10 3x = 10004x = 250ln(2x + 3) − 2 ln 7 = 1.5ln 2x + 37 2 = 1.52x + 3= e 1.5492x + 3 = 49e 1.52x = 49e 1.5 − 3x = 49e1.5 − 32x ≈ 108.3Solving exponential equations (with logs)Hint: The key step is to take the log of both sides. It may help to rearrange it first though.In all the solutions, I simplify fully in log form before writing the approximate answer as adecimal number. You may not have to do that for class.10 x = 500log(10 x ) = log(500)x log 10 = log(5) + log(100)x = log 5 + 2x ≈ 2.699e −2x − 1 = 4e −2x = 5ln(e −2x ) = ln(5)−2x ln(e) = ln 5−2x = ln 5x = − ln 52x ≈ −0.80472(5 x+2 ) = 3005 x+2 = 3002ln(5 x+2 ) = ln150(x + 2) ln(5) = ln150x + 2 = ln150ln 5x = ln150ln 5 − 2x ≈ 1.11www.durhamcollege.ca/sals Student Services Building (SSB), Room 204905.721.2000 ext. 2491This document last updated: 7/5/2012