Lesson 3 - Solve with Substitution and Log Properties
Download
Report
Transcript Lesson 3 - Solve with Substitution and Log Properties
Solve with
Substitution
Solve by Using Substitution
2
Let y 2
2x
62 8 0
x
y2 6y 8 0
y 2 y 4 0
y2
y4
x
y 2x
y 2x
2 2x
4 2x
21 2 x
x 1
22 2 x
x2
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
x3
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
x2
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
x2
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