Lesson 3 - Solve with Substitution and Log Properties

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Transcript Lesson 3 - Solve with Substitution and Log Properties

Solve with
Substitution
Solve by Using Substitution
2
Let y  2
2x
 62 8  0
x
y2  6y  8  0
 y  2 y  4  0
y2
y4
x
y  2x
y  2x
2  2x
4  2x
21  2 x
x 1
22  2 x
x2
Solve by Using Substitution
5  130  5  625  0
2x
Let y  5
x
x
y 2  130 y  625  0
 y  5 y  125  0
y 5
y  125
y  5x
y  5x
5  5x
125  5 x
51  5 x
x 1
53  5 x
x3
Solve by Using Substitution
3  12  3  27  0
2x
Let y  3
x
y 2  12 y  27  0
 y  3 y  9  0
y 3 y 9
x
y  3x
y  3x
3  3x
9  3x
31  3 x
x 1
32  3 x
x2
Logarithmic
Properties
Properties of Logarithms
log b ( mn)  log b m  log b n
m
log b    log b m  log b n
n
log b m p  p log b m
Simplify:
(Condense)
log8 12  log8 4
12
 log 8
4
 log 8 3
Simplify:
logb u  logb v  logb uw
Step 1 : log b uv  log b uw
uv
Step 2 : log b
uw
v
Final Answer : log b
w
Simplify:
log 4 18  log 4 6
18
 log 4
6
 log 4 3
Simplify:
logb 2a  logb b  logb bc
2a
 log b
 log b bc
b
 2a bc 
 log b   
 b 1
 log b 2ac
Simplify:
log 4 x  log 3 y  log y
4x
 log
 log y
3y
 4 xy 

 log 
 3y 
 4x 
 log  
 3 
Simplify:
log xy  log 2 x  log y
 log  xy  2 x  y 

 log 2 x y
2
2

Simplify:
3log 4 x  2log 4 y
 log 4 x  log 4 y
3
x
 log 4  2
y
3



2
7 log3 y  4log3 x
Simplify:
 log 3 y  log 3 x
7
4
7
y
 log 3 4
x
TAKE NOTE :
Simplify : 2 log  xy
 log  xy
2
 log x 2 y 2
Simplify:
2  4 log3 y
 2  log 3 y
4
Simplify and
Evaluate if
Possible:
log3 45  log3 5
45
 log 3
5
 log 3 9
Keep Going .....
x  log 3 9
3 9
x2
x
Expand:
log3 8xy
 log 3 8  log 3 xy
 log 3 8  log 3 x  log 3 y
Expand:
 mn 
log b 

 k 
 log b mn  log b k
 log b m  log b n  log b k
Expand:
 x 
log 4  
 32 
 log 4 x  log 4 32
You can find log 4 32 :
x  log 4 32
4 x  32
5
 log 4 x 
2
OR
5
   log 4 x
2
2 2 x  25
2x  5
5
x
2
Expand:
 k 
log b 

 mn 
 log b k  log b mn
 log b k  (log b m  log b n)
 log b k  log b m  log b n