Assignment, pencil, red pen,
highlighter, notebook, calculator
Solve for x.
1) log23x 1  4


24  3x 1 +1
16  3x 1 +1
15 3x
x = 5 +2
x3  27
3
+1
x  27 +1
3
3
x=3
total:
14


3) log5 x 2  35  log5 2x 
2) logx 27  3
+2
x 2  35  2x +1
x 2  2x  35  0
x  7x  5  0 +1
x – 7 = 0 or x +5 = 0 +1
x = –5 +1
x = 7 +1
Check!!!
x = 7 +1
Rule:
1)
Example:
b b 
x y
bx+y
2)
bx
x–y
b

by
3)
b 
x y
 bxy
Keyword:
3+5 =28
2
22 
Product
38
8–5
3

3
=
3
35
Quotient
3 5
4 
2 3
 42(3) = 46
Power
Argument
Log 2 8  3
Base
Exponent
Complete the calculations for Part 1a on your calculator.
Be sure to locate the LOG function on your calculator.
LOG button
Note: the LOG function
on your calculator is
base 10, that is:
log( ) means log10( )
log 5 = _________
0.70
log 6 = _________
0.78
log 30 = 1.48
_________
0.85
log 7 = ________
0.30
log 2 = ________
1.15
log 14 = ________
1.08
log 12 = _________
0.48
log 3 = _________
1.56
log 36 = _________
multiply the
When you ________
add
arguments, you ______
the
exponents.
Product Property of Logarithms
The __________
x • y
log x + log y = log (________)
Conversely: log (x ∙ y) = log (x) ____
+ log (y)
log b x  log b y  log b (x  y)
1c) Practice: Use the property to rewrite each expression as a
single logarithm.
i) log2 9  log2 5
log 2 (9  5)
log 2 (45)
1
 log 6 10
2
1

log 6  10 
2

ii) log 6
iii) loga 3b  loga 8c 
log a 3b  8c 
log a 24bc 
log 6 5 
1d) Practice: Express in expanded form.
i) loga xz
log a x  log a z
ii) log4 3y
log 4 3  log 4 y
iii) log7x
log7  logx
1.90
log 80 = _________
0.90
log 8 = __________
1.0
log 10 = _________
1.75
log 56 = _________
0.90
log 8 = _________
0.85
log 7 = __________
log 36 = _________
1.56
0.60
log 4 = __________
log 9 = __________
0.95
divide
When you _______
the
subtract the
arguments, you ________
exponents.
The __________
Quotient Property of Logarithms
x
log x – log y = log (
)
y
x
– log (y)
Conversely: log  y  = log (x) ____


x
logb x  logb y  logb  
y
2c) Practice: Use the property to rewrite each expression as
a single logarithm.
i) log710  log7 3 ii) log512  log5 4
iii) logx y  logx z
 10 
log 7  
 3 
 12 
log 5  
 4 
log 5 3 
y
log x  
z
2d) Practice: Express in expanded form.
a
8
x
i) logb
ii) log 2
iii) log 9
c
x
4
logba  log b c
log 2 x  log 2 4
log 2 x  2
log 9 8  log 9 x
Use the
product rule!
log (52) = log ( ________)
log (82) = log ( ________)
5 ∙ 5
8 ∙ 8
8
5
5 + log (____)
8 + log (__)
= log (____)
= log (__)
2 log 8
2 log 5
= __________________
= ______________
Write log 63 as the sum of three logs, Let’s use a calculator:
1.91
then simplify:
Find: log (34) =________
6∙6∙6
1.91
log (63) = log ( ________)
Find: 4 log 3 = _______
6
6 + log (__)
6 + log (__)
= log (__)
3 log 6
= __________________
Power Property of Logarithms
The ______
log (ab) = (___)log
a
b
ab
Conversely: b log a = log (____)
b log a  log ab
3c) Practice: Use the property to rewrite each expression.
i) log4 x2
2log 4 x
ii) log 5
1
y3
1
log 5 y
3
iii) log2 84
4log 2 8
4(3)
12
3d) Practice: Rewrite without the coefficients, then simplify.
1
i) 5 logb x
ii) log 3 25
iii) 2 log 6 7
2
logb x
5
1
25 2
log3
log 3 5
log 6 7 2
log 6 49
MIXED PRACTICE
1. Rewrite each expression as a single logarithm. Simplify, if
possible.
a) log 5 + log 7
log (5 • 7)
log 35
b) log 15 – log 5
 15 
log 
 5 
log 3
c) log 9 – log 3
9
log 
3
log 3
d) log2 8  log2x
e) log3 x  2 log3 y
f) log5 x  log5 9
x
log 5  
9
log 2 8x 
log 3 x  log 3 y 2
 
log 3 xy 2
MIXED PRACTICE
1. Rewrite each expression as a single logarithm. Simplify, if
possible.
g) log3 y  4 log3 t
log 3 y  log 3 t
4
 
log 3 yt 4
j) 3 log4 y  2 log4z
log 4 y 3  log 4 z 2
 y3 
log 4  2 
z 
h) 7 log p  log q
i) 5 log2m  log2p
log p 7  log q
log 2m 5  log 2p
log p 7 q
 m5 

log 2 
 p 


k) log4  log 5  log 6
1
1
l) log 2 4  log 2 x
2
3
1
1
 
4
log   log 6
5
4 
log  6 
5 
 24 
log 
 5 
log 2 4 2  log 2 x 3
log 2 2  log 2 3 x
 2 
log 2  3 
 x
2. Express in expanded form.
a) logdac
log da  log dc
x
d) logb
10
log b x  log b 10
b) log6 9z
log 6 9  log 6 z
log 6 3 2  log 6 z
2log 6 3  log 6 z
e) log8 7x 2
log 8 7  log 8 x 2
log 8 7  2log 8 x
y
c) log a
c
log a y  log a c
y2
f) log3
4
log 3 y 2  log 3 4
2log 3 y  log 3 4
3. Solve the equations. Check your solutions.
a) log3 5  log3 x  log310 To solve problems such as this, we
log3 5x   log310
5x  10
x2
need to use the product property to
put the logs on the left side
together.
Now the problem is written:
log ( ) = log ( )
Use the property of equality to
set the arguments equal, then
solve.
We are going to work on the problems in the first column.
Skip to part (c) now.
3. Solve the equations. Check your solutions.
c) log 16  log2t   log 2 Use the quotient property to
combine the left side.
 16 
log    log 2
 2t 
16
2
2t
16 2

2t 1
4t  16
t4
Now the problem is written:
log ( ) = log ( )
Use the property of equality to
set the arguments equal, then
solve.
3. Solve the equations. Check your solutions.
e) logc x  2 logc 4  logc 2 Use the power property to
move the coefficient.
logc x  logc 42  logc 2
Use the quotient property to
combine the right side.
 16 
logc x  logc  
 2
Now the problem is written:
log ( ) = log ( )
logc x  logc 8
x 8
Use the property of equality to
set the arguments equal, then
solve.
3. Solve the equations. Check your solutions.
g) log 2 n 
1
1
log 216  log 2 49
4
2
log2 n  log2
1
16 4
 log2
log2 n  log2 2  log2 7
log2 n  log2 2  7
n  27
n  14
Use the power property to
move the coefficient.
1
49 2
Use the product property to
combine the right side.
Now the problem is written:
log ( ) = log ( )
Use the property of equality to
set the arguments equal, then
solve.