Rules of Logs1: 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 > -33: 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 87: Log (x) – log (y) = log (x/y) Ex: log35 – log32 log3 (5/2)8: Log (1) = 0 Ex: log3(1) 09: Log3(x) = log3 (y) then x = y Ex: log(4) = log(x+1) 4 = x + 1 x = 310: Loga(x) = b then ab = x Ex: log3(x) = 2 32 = x11: (1 + 1/x)x =~ 2.178281828459045 = e12: Logex = ln (x) Ln and e undo each other. Ln follows log rules13: ex =b then ln (b) = x Ex: ex+3= 5 x+3 = ln 5 x = ln 5 – 314: a = ln(x) then ea = x Ex: 5 = ln(x+3) e5 = x + 315: Ln(x) = ln(y) then x = y Ex: ln(x+1) = ln 5 x + 1 =5 x = 416: ex = ey then x = y Ex: ex = e5 x = 517: Ln em = m Ex: ln (e)4 418: elnm = m Ex: eln4 419: logam can be written as log(m)log(a) Ex: log38 on calculator log (8)/log(3)

xa xa log 95 9log5

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= 81-3/4 = x {

1. log3 812. Log4(1/64)3. Log .00014. Log2x = -35. Logmm 6. log22x = 27. Log8138. 3log

36

9. Log64x=-2/3

1. 42. -33. -44. 1/85. 16. 4847. ¼8. 69. 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 = 1.51.5

10. =m 27 = m

33 = 3(1/2)m

3 = .5m 3/.5 (mult by 10/10) 30/56

11. Log64128 log 64128 = m 128 = 64m

27 = 26m

7 = 6m 7/6 = m 7/6

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 themA number in front of log becomes the exponent.Minus logs, minus logs, divide themLog 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

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

Expanding Logs. No exponents! Bottom logs get ‘-’

Expand the following log: log log x4 – log 3 – log m2

4logx – log3 – 2logm

Expand the following log: log 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 Log45x = log460 5x = 60 x = 12CHECK answer!{12}

24) log2(x+4) – log2(x-3)= 3

33

4log2

x

x

323

4

x

x

83

4

x

x

8(x-3) = x+48x-24=x + 47x = 28 x=4 CHECK!{4}

18) 3 log82 – log84 = log8b

log823 – log84 = log8b

log8(8/4) = log8bLog82 = log8b 2 = b

Problems taken from Glencoe Algebra II workbook 10.3

11) log1027 = 3 log10 x 12) 3log74=2 log7b

Log1027 = log10x3

27 = x3

3 = x{3}

log743 = log7b2

log764 =log7b2

64 = b2

+ 8 = b{8}

Problems taken from Glencoe Algebra II workbook 10.3

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}

21) log3d + log33 = 3 23) log2s + 2 log25 =0

Log3 (d3) = 3 3d = 33

3d = 27 d = 9

Log2s + log252 = 0Log2(s25) = 0 25s = 20

25s = 1 s = 1/25

Problems taken from Glencoe Algebra II workbook 10.3

20) log10x + log10(3x – 5) = 219) log4x+ log4(2x – 3) = log42

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}

Ln and e (They “undo” each other)a=eb ln a = b

Ex: Rewrite z = ln 2x into exponential formez = 2x

Ex: Rewrite 5 = e2x into natural log formLn 5 = 2x

13. Rewrite c = ln x into exponential formec = x

14. Rewrite e5c = ma into5c = ln (ma)

(1 + 1/x)x as x gets larger….. 2.718281828459045 ln is loge

Solving ln and e equations (follow rules)

ex = ey x = y ln x = ln y x =yex = # x = ln(#) ln(x) = a x = ea

Ex: e3x *e4 = e2x/e4

e3x+4 = e2x-4

3x+ 4 = 2x – 4

X=-8

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 5e2x+1 +7 = 32 5e2x+1 = 25 e2x+1 = 5 2x + 1 = ln 5 2x = ln (5) – 1 x =

Ex: ln(x) – 32 > - 2ln(5) <all ln on one side>

ln(x) + ln(5)2> 32ln 25x > 32 25x > e32

x >

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 + 322x2 – 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 Circumstancee2x + 3ex = 4e2x + 3ex – 4=0(ex + 4)(ex - 1)=0ex = -4 ex = 1x = 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 {}

Typing Logs into CalculatorLogab =

Log381 = log123 =

21. Log515 log 15 log 5

22. Log168 log 8 log 16

Finding Inverses of Exponential and Log Equations

Ex: Y = 3x + 2 x = 3y + 2 x-2 = 3y

log3(x-2) = y f-1(x) = log3(x-2)

1. Switch x and y.2. Solve for the new y.3. Use looping method to write it as a log or exponential

ay = x y = logax

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

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

25: y = e5x + 1 x = e5y + 1 x-1 = e5y

ln(x-1) = 5y= f-1(x)

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 logsLoga2=.34 loga5=.68Find loga10 loga2/5 loga8a

loga 1/25

Loga(2*5) = loga2+ loga5 = .34 + .68 = 1.02Loga(2/5) = loga2 – loga5 = .34 - .68 = -.34Loga(2*2*2*a)=loga2+loga2+loga2+logaa =.34 + .34 + .34 + 1 = 2.02Loga(1/(5*5) =loga1-loga5 – loga5 = 0 - .68 - .68 = -1.36

Loga6=.26 loga12=.4Find loga72 loga0.5 loga36a

loga 3

Loga(6*12) = loga6 + loga12 = .26 + .4 = .66Loga(6/12) = loga6 – loga12 = .26-.4 = -.14Loga(6*6*a) = loga6 + loga6 + logaa = .26+.26+1 = 1.52

Loga(6*6/12) = loga6+ loga6 – loga12 = .26+.26-.4 = .12

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 2X + 1 = X= - 1

Ex: 5x+1 = 2x

Log 5x+1 = log 2x

(x+1)log5 = xlog 2xlog 5 + 1 log 5 =x log 2xlog 5 – x log 2 = log 5x(log 5 – log 2) = log 5x =

Ex: 32x+1 = 2x+4

Log 32x+1 = log 2x+4

(2x+1)log3 = (x+4)log 22xlog 3 + 1log 3 =xlog2+4log2

2xlog3–xlog 2 = 4log2-log3X(2log3 –log 2) =4log2-log3X =

29. 4x+3 = 2 Log 4x+3 = log 2(x+3)log4 = log 2X + 3 = X= - 3

30. 6x-1 = 2x

Log 6x-1 = log 2x

(x-1)log6 = xlog 2xlog 6 - 1 log 5 =x log 2xlog 6 – x log 2 = log 6x(log 6 – log 2) = log 6x =

31. 83x+1 = 2x+4

Log 83x+1 = log 2x+4

(3x+1)log8 = (x+4)log 23xlog 8 + 1log 8 =xlog2+4log2

3xlog8–xlog 2 = 4log2-log8X(3log8 –log 2) =4log2-log8X = =