Rules of Logs 1: A log with no base has a base of 10 Ex: log 100 = 2  log10 100 = 2  100 = 102 2: Domain of logs log (~)  ~ > 0 Ex: log (x+3)  x+3 > 0  x > -3 3: Log a Ax = x Ex: log334  4 5: Log (x) + log (y) = log (xy) Ex: log (3) + log (5) = log(15) 6: b log(x) = log xb Ex: 3 log 2  log 23  log 8 7: Log (x) – log (y) = log (x/y) Ex: log35 – log32  log3 (5/2) 8: Log (1) = 0 Ex: log3(1)  0 9: Log3(x) = log3 (y) then x = y Ex: log(4) = log(x+1)  4 = x + 1  x = 3 10: Loga(x) = b then ab = x Ex: log3(x) = 2  32 = x 11: (1 + 1/x)x =~ 2.178281828459045 = e 12: Logex = ln (x) Ln and e undo each other. Ln follows log rules 13: ex =b then ln (b) = x Ex: ex+3= 5  x+3 = ln 5  x = ln 5 – 3 14: a = ln(x) then ea = x Ex: 5 = ln(x+3)  e5 = x + 3 15: Ln(x) = ln(y) then x = y Ex: ln(x+1) = ln 5  x + 1 =5  x = 4 16: ex = ey then x = y Ex: ex = e5  x = 5 17: Ln em = m Ex: ln (e)4  4 18: elnm = m Ex: eln4  4 19: logam can be written as log(m)log(a) Ex: log38 on calculator  log (8)/log(3)

Simplifying Logs Logab is written as ‘a’ to what power is ‘b’ Ex: log3 9 is read as 3 to what power is 9 2 Ex: log .001 10 to what power is .001  -3 Ex: log2(1/128) 2 to what power is 1/128  -7 Ex: log255 25 to what power is 5  ½ Ex: log4x = 2 4 to the 2 power is x  16 Ex: log 1 10 to what power is 1  0 Ex: log55-3 -3 because 5’s cancel Ex: 10log4 4 because 10 and log10cancel Ex: log81x= −3 4  81-3/4 = x  1 81 3/4  1 4 81 3  1 3 3 { 1 27 }

log3 81 Log4(1/64) 3. Log .0001 4. Log2x = -3 5. Logmm log22x = 2 Log813 3log36 Log64x=-2/3 1. 4 -3 -4 4. 1/8 5. 1 484 ¼ 6 64-2/3 = 1 64 2/3  1/16

Occassionally, you might want to put a “m” into the problem to make it easier to simplify Log25125 log 25125 = m 125 = 25m 53 = 52m 3 = 2m m = 1.5 1.5 10. log 3 27 =m 27 = 3 m 33 = 3(1/2)m 3 = .5m 3/.5 (mult by 10/10) 30/5 6 11. Log64128 log 64128 = m 128 = 64m 27 = 26m 7 = 6m 7/6 = m 7/6

Adding logs, Adding logs, multiply them a3a5 = a3+5 loga3 + loga5 = loga(3*5) (a3)5 = a3*5 5loga3 = loga35 a5 = a5-3 a3 loga5 – loga3 = loga (5/3) a0 = 1 Loga1=0 ax is always positive Loga(~) ~ is always positive!! Adding logs, Adding logs, multiply them A number in front of log becomes the exponent. Minus logs, minus logs, divide them Log of 1 is zero, and can’t take log of negative.

Condensing Logs -logs go on bottom! Write as one log: 3logx + 2log y – 1/2log 9 – 3logm logx3 + log y2 – log 91/2 – log m3 logx3 + log y2 – log 3 – log m3 log 𝑥 3 𝑦 2 3 𝑚 3 Write as one log: -4logx - 2log y – 1/3log 64 – 3logm -logx4 - log y2 – log 641/3 – log m3 -logx4 - log y2 – log 4 – log m3 log 1 4𝑥 4 𝑦 2 𝑚 3

Expanding Logs. No exponents! Bottom logs get ‘-’ Expand the following log: log 𝑥 4 3 𝑚 2 log x4 – log 3 – log m2 4logx – log3 – 2logm Expand the following log: log 𝑦 3 3 𝑚 2 log y3 – log 3 – log m2 3logy – log3 – 2logm

Recall that if logx5 = logxy then 5=y <Log=Log> logx25 = 2 then 25 = x2 <Log = #> Condense logs to just ONE LOG on each side and then solve by log=log or log = # 13) log45 + log4x = log460 24) log2(x+4) – log2(x-3)= 3 Log45x = log460 5x = 60 x = 12 CHECK answer! {12} 18) 3 log82 – log84 = log8b log823 – log84 = log8b 8(x-3) = x+4 8x-24=x + 4 7x = 28 x=4 CHECK! {4} log8(8/4) = log8b Log82 = log8b 2 = b Problems taken from Glencoe Algebra II workbook 10.3

12) 3log74=2 log7b 11) log1027 = 3 log10 x log743 = log7b2 {8} Log1027 = log10x3 27 = x3 3 = x {3} 16) Log2q – log23 = log27 15) log5y-log58=log51 Log2 (q/3) = log27 q/3 = 7 q = 21 {21} Log5 y/8 = log5 1 y/8 = 1 y = 8 {8} Problems taken from Glencoe Algebra II workbook 10.3

21) log3d + log33 = 3 Log2s + log252 = 0 Log2(s25) = 0 25s = 20 19) log4x+ log4(2x – 3) = log42 20) log10x + log10(3x – 5) = 2 Log4 (x(2x -3)) = log4(2) 2x2 – 3x = 2 2x2 – 3x – 2 = 0 (2x + 1)(x – 2) = 0 2x + 1= 0 x-2=0 x = -1/2 x =2 {2} Log10(x(3x-5)) = 2 3x2 – 5x = 102 3x2 – 5x = 100 3x2 – 5x – 100 = 0 (3x -20 )(x+5 )=0 x = 20/3 x = -5 {20/3} Problems taken from Glencoe Algebra II workbook 10.3

Ln and e (They “undo” each other) a=eb  ln a = b (1 + 1/x)x as x gets larger….. 2.718281828459045 ln is loge Ex: Rewrite z = ln 2x into exponential form ez = 2x Ex: Rewrite 5 = e2x into natural log form Ln 5 = 2x 13. Rewrite c = ln x into exponential form ec = x 14. Rewrite e5c = ma into 5c = ln (ma)

Solving ln and e equations (follow rules) ex = ey  x = y ln x = ln y  x =y ex = #  x = ln(#) ln(x) = a  x = ea Ex: ln(x) + 2ln(4)= ln(32) ln(x) + ln(4)2 = ln 32 ln x + ln 16 = ln 32 ln 16x = ln 32 16x = 32  x = 2 Ex: e3x *e4 = e2x/e4 e3x+4 = e2x-4 3x+ 4 = 2x – 4 X=-8 Ex 5e2x+1 +7 = 32 5e2x+1 = 25 e2x+1 = 5 2x + 1 = ln 5 2x = ln (5) – 1 x = ln 5 −1 2 Ex: ln(x) – 32 > - 2ln(5) <all ln on one side> ln(x) + ln(5)2> 32 ln 25x > 32 25x > e32 x > 𝑒 32 25

ex = #  x = ln(#) ln(x) = a  x = ea ex = ey  x = y ln x = ln y  x = ex = #  x = ln(#) ln(x) = a  x = ea 15. 25e2x+1 -7 = -32 25e2x+1 = -25 e2x+1 = -1 2x + 1 = ln (-1) Impossible No Solution 16. 2ln(x) + ln (2) = ln (12x+32) ln(x2) + ln (2) = ln (12x + 32) ln (2x2 ) = ln (12x + 32) 2x2 = 12x + 32 2x2 – 12x – 32 = 0 2(x2 – 6x – 16) (x-8)(x+2)  x=8, x= -2 {8} 17. ln(8) – ln(2) + 5= -ln x ln (8) – ln(2) + ln x = -5 ln (8x/2) = -5 ln(4x) = -5 4x = e-5 x = e-5/4 18. e3x * e = e6 e3x+1 = e6 3x + 1 = 6 3x = 5 x=5/3

Special Factoring Circumstance e2x + 3ex = 4 e2x + 3ex – 4=0 (ex + 4)(ex - 1)=0 ex = -4 ex = 1 x = ln (-4) x = ln(1) {ln 1} only e2x is 2times ex (x-4)(x+1)=x2+3x-4 20. e4x + 5e2x = 6 e4x + 5e2x – 6=0 (e2x + 6)(e2x - 1)=0 e2x = -6 e2x = 1 2x = ln (-6) 2x = ln(1) x = ln(-6)/2 x= ln(1)/2 { ln⁡(1) 2 }

Typing Logs into Calculator Logab = 𝑙𝑜𝑔𝑏 𝑙𝑜𝑔𝑎 Log381 = 𝑙𝑜𝑔81 𝑙𝑜𝑔3 log123 = 𝑙𝑜𝑔3 𝑙𝑜𝑔12 21. Log515 log 15 log 5 22. Log168 log 8 log 16

Finding Inverses of Exponential and Log Equations Switch x and y. Solve for the new y. Use looping method to write it as a log or exponential ay = x  y = logax Ex: Y = 3x + 2 x = 3y + 2 x-2 = 3y log3(x-2) = y f-1(x) = log3(x-2) Ex: y= log3(x-2) x= log3(y-2) 3x = y - 2 3x + 2 = y 3x + 2 = f-1(x) Ex: y= ln(x-2) x= ln(y-2) ex = y - 2 ex + 2 = y f-1(x)=ex + 2

25: y = e5x + 1 23: f(x)=5x - 1 24: f(x) = 10x-1 x = e5y + 1 ln(x-1) = 5y ln⁡(𝑥−1) 5 = f-1(x) 23: f(x)=5x - 1 x = 5y - 1 x+1 = 5y log5(x+1) = y f-1(x) = log5(x+1) 24: f(x) = 10x-1 x = 10y-1 log(x) = y-1 f-1(x)= log(x)+1 26: y= log4(x-2)+1 x= log4(y-2)+1 x-1=log4(y-2) 4x-1 = y - 2 4x-1 + 2= y 4x-1 + 2 = f-1(x) 27: y= log(x-3)+1 x = log(y-3)+1 x-1 = log(y-3) 10x-1 = y-3 10x-1 = y - 3 10x-1 + 3= y 10x-1 + 3 = f-1(x) 28: y= ln(x+3)-1 x = ln (y+3) - 1 x+ 1= ln(y+3) ex+1 = y +3 ex+1 - 3= y f-1(x) = ex+1 - 3

Finding value of logs given some logs Loga2=.34 loga5=.68 Find loga10 loga2/5 loga8a loga 1/25 Loga(2*5) = loga2+ loga5 = .34 + .68 = 1.02 Loga(2/5) = loga2 – loga5 = .34 - .68 = -.34 Loga(2*2*2*a)=loga2+loga2+loga2+logaa =.34 + .34 + .34 + 1 = 2.02 Loga(1/(5*5) =loga1-loga5 – loga5 = 0 - .68 - .68 = -1.36 Loga6=.26 loga12=.4 Find loga72 loga0.5 loga36a loga 3 Loga(6*12) = loga6 + loga12 = .26 + .4 = .66 Loga(6/12) = loga6 – loga12 = .26-.4 = -.14 Loga(6*6*a) = loga6 + loga6 + logaa = .26+.26+1 = 1.52 Loga(6*6/12) = loga6+ loga6 – loga12 = .26+.26-.4 = .12

X = (4𝑙𝑜𝑔2−𝑙𝑜𝑔3) (2𝑙𝑜𝑔3 −𝑙𝑜𝑔2) Solving Exponential Equations (when exponent is unknown then put logs to both sides) Put all x’s to one side, factor it out and then divide! Ex: 5x+1 = 2 Log 5x+1 = log 2 (x+1)log5 = log 2 X + 1 = 𝑙𝑜𝑔2 𝑙𝑜𝑔5 X= 𝑙𝑜𝑔2 𝑙𝑜𝑔5 - 1 Ex: 5x+1 = 2x Log 5x+1 = log 2x (x+1)log5 = xlog 2 xlog 5 + 1 log 5 =x log 2 xlog 5 – x log 2 = log 5 x(log 5 – log 2) = log 5 x = 𝑙𝑜𝑔5 (𝑙𝑜𝑔5 −𝑙𝑜𝑔2) Ex: 32x+1 = 2x+4 Log 32x+1 = log 2x+4 (2x+1)log3 = (x+4)log 2 2xlog 3 + 1log 3 =xlog2+4log2 2xlog3–xlog 2 = 4log2-log3 X(2log3 –log 2) =4log2-log3 X = (4𝑙𝑜𝑔2−𝑙𝑜𝑔3) (2𝑙𝑜𝑔3 −𝑙𝑜𝑔2) 30. 6x-1 = 2x Log 6x-1 = log 2x (x-1)log6 = xlog 2 xlog 6 - 1 log 5 =x log 2 xlog 6 – x log 2 = log 6 x(log 6 – log 2) = log 6 x = 𝑙𝑜𝑔6 (𝑙𝑜𝑔6 −𝑙𝑜𝑔2) 31. 83x+1 = 2x+4 Log 83x+1 = log 2x+4 (3x+1)log8 = (x+4)log 2 3xlog 8 + 1log 8 =xlog2+4log2 3xlog8–xlog 2 = 4log2-log8 X(3log8 –log 2) =4log2-log8 X = (4𝑙𝑜𝑔2−𝑙𝑜𝑔8) (3𝑙𝑜𝑔8 −𝑙𝑜𝑔2) = 29. 4x+3 = 2 Log 4x+3 = log 2 (x+3)log4 = log 2 X + 3 = 𝑙𝑜𝑔2 𝑙𝑜𝑔4 X= 𝑙𝑜𝑔2 𝑙𝑜𝑔4 - 3