Rules of Logs
1: A log with no base has a base of 10
2: Domain of logs log (~)  ~ > 0
3: Log a Ax = x
loga x
a

x
5: Log (x) + log (y) = log (xy)
Ex: log 100 = 2  log10 100 = 2  100 = 102
Ex: log (x+3)  x+3 > 0  x > -3
Ex: log334  4
59
5Ex:loglog

9
(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)
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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:
−3
log81x= 
4
81-3/4
=x
1
1
1
1
 4 3 3{ }
3/4
81
3
27
81
1. log3 81
2. Log4(1/64)
3. Log .0001
4. Log2x = -3
5. Logmm
6. log22x = 2
7. Log813
8. 3log 6
9. Log64x=-2/3
3
1.
2.
3.
4.
5.
6.
7.
8.
9.
4
-3
-4
1/8
1
484
¼
6
64-2/3 =
1
 1/16
2/3
64
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
6
10. log 3 27 =m
11. Log64128
27 = 3 m
log 64128 = m
33 = 3(1/2)m
128 = 64m
3 = .5m
27 = 26m
3/.5 (mult by 10/10)
7 = 6m
30/5
7/6 = m
7/6
a3a5 = a3+5
(a3)5 = a3*5
loga3 + loga5 = loga(3*5)
5loga3 = loga35
a5 = a5-3
a3
a0 = 1
loga5 – loga3 = loga (5/3)
ax is always positive
Loga1=0
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
𝑥3𝑦2
3𝑚3
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
1
log 4 2 3
4𝑥 𝑦 𝑚
Expanding Logs.
No exponents! Bottom logs get ‘-’
𝑥4
3𝑚2
Expand the following log: log
log x4 – log 3 – log m2
4logx – log3 – 2logm
𝑦3
3𝑚2
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 = 12
CHECK answer!
{12}
24) log2(x+4) – log2(x-3)= 3
18) 3 log82 – log84 = log8b
log823 – log84 = log8b
log8(8/4) = log8b
Log82 = log8b
2=b
Problems taken from Glencoe Algebra II workbook 10.3
x4
log 2
3
x3
x4
 23
x3
x4
8
x3
8(x-3) = x+4
8x-24=x + 4
7x = 28
x=4 CHECK!
{4}
11) log1027 = 3 log10 x
Log1027 = log10x3
27 = x3
3=x
{3}
16) Log2q – log23 = log27
12) 3log74=2 log7b
log743 = log7b2
log764 =log7b2
64 = b2
+8=b
{8}
15) log5y-log58=log51
Log2 (q/3) = log27
q/3 = 7
q = 21
{21}
Problems taken from Glencoe Algebra II workbook 10.3
Log5 y/8 = log5 1
y/8 = 1
y=8
{8}
21) log3d + log33 = 3
Log3 (d3) = 3
3d = 33
3d = 27
d=9
19) log4x+ log4(2x – 3) = log42
23) log2s
+ 2 log25 =0
Log2s + log252 = 0
Log2(s25) = 0
25s = 20
25s = 1
s = 1/25
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}
Problems taken from Glencoe Algebra II workbook 10.3
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
(1 + 1/x)x as x gets larger…..
Ex: Rewrite z = ln 2x
into exponential form
ez = 2x
13. Rewrite c = ln x into
exponential form
ec = x
2.718281828459045 ln is loge
Ex: Rewrite 5 = e2x into
natural log form
Ln 5 = 2x
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:
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) + 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: ln(x) – 32 > - 2ln(5)
ln(x) + ln(5)2> 32
ln 25x > 32
25x > e32
x
𝑒 32
>
25
<all ln on one side>
ex = ey
ex = #
x=y
 x = ln(#)
15. 25e2x+1 -7 = -32
25e2x+1 = -25
e2x+1 = -1
2x + 1 = ln (-1)
Impossible
No Solution
ln x = ln y  x =
ln(x) = a  x = ea
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 is 2times ex
e2x + 3ex – 4=0
(x-4)(x+1)=x2+3x-4
x
x
(e + 4)(e - 1)=0
ex = -4
ex = 1
20. e4x + 5e2x = 6
x = ln (-4) x = ln(1)
e4x + 5e2x – 6=0
{ln 1} only
(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
𝑙𝑜𝑔𝑏
Logab =
𝑙𝑜𝑔𝑎
𝑙𝑜𝑔81
Log381 =
𝑙𝑜𝑔3
21. Log515
log 15
log 5
log123
𝑙𝑜𝑔3
=
𝑙𝑜𝑔12
22. Log168
log 8
log 16
Finding Inverses of Exponential and Log Equations
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 = 3x + 2 Ex:
x = 3y + 2
x-2 = 3y
log3(x-2) = y
f-1(x) = log3(x-2)
y= log3(x-2) Ex: y= ln(x-2)
x= log3(y-2)
x= ln(y-2)
3x = y - 2
ex = y - 2
3x + 2 = y
ex + 2 = y
3x + 2 = f-1(x)
f-1(x)=ex + 2
5x + 1
25:
y
=
e
23:
- 1 24: f(x) =
5y + 1
y
x
=
e
y-1
x=5 -1
x = 10
5y
y
x-1
=
e
x+1 = 5
log(x) = y-1
log5(x+1) = y
f-1(x)= log(x)+1 ln(x-1) = 5y
ln(𝑥−1) -1
-1
f (x) = log5(x+1)
= f (x)
f(x)=5x
10x-1
5
26: y= log4(x-2)+1 27:
x= log4(y-2)+1
x-1=log4(y-2)
4x-1 = y - 2
4x-1 + 2= y
4x-1 + 2 = f-1(x)
y= log(x-3)+1 28: y= ln(x+3)-1
x = ln (y+3) - 1
x = log(y-3)+1
x+ 1= ln(y+3)
x-1 = log(y-3)
ex+1 = y +3
10x-1 = y-3
ex+1 - 3= y
10x-1 = y - 3
f-1(x) = ex+1 - 3
10x-1 + 3= y
10x-1 + 3 = f-1(x)
Finding value of logs given some logs
Loga2=.34
Find loga10
loga2/5
loga8a
loga5=.68
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
loga 1/25
Loga6=.26
Find loga72
loga0.5
loga36a
loga 3
loga12=.4
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
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 = 2x
Ex: 5x+1 = 2
Log 5x+1 = log 2 Log 5x+1 = log 2x
(x+1)log5 = log 2 (x+1)log5 = xlog 2
𝑙𝑜𝑔2
xlog 5 + 1 log 5 =x log 2
X+1=
𝑙𝑜𝑔5
xlog 5 – x log 2 = log 5
𝑙𝑜𝑔2
X=
-1
x(log 5 – log 2) = log 5
𝑙𝑜𝑔5
𝑙𝑜𝑔5
x=
(𝑙𝑜𝑔5 −𝑙𝑜𝑔2)
30. 6x-1 = 2x
29. 4x+3 = 2
Log 6x-1 = log 2x
Log 4x+3 = log 2 (x-1)log6 = xlog 2
(x+3)log4 = log 2 xlog 6 - 1 log 5 =x log 2
𝑙𝑜𝑔2
xlog 6 – x log 2 = log 6
X+3=
𝑙𝑜𝑔4
x(log 6 – log 2) = log 6
𝑙𝑜𝑔2
𝑙𝑜𝑔6
X=
-3
x=
𝑙𝑜𝑔4
(𝑙𝑜𝑔6 −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)
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)